CBSE Class 12 Mathematics Application Of Derivative Worksheet Set 01

Points to Remember

• Let \( f \) be a function. A point \( c \) in the domain of \( f \) at which either the derivative \( f'(c) = 0 \) or the function \( f \) is not differentiable is called a critical point of \( f \).
• First Derivative Test: Let \( f \) be a continuous function defined on an open interval \( I \) and let \( c \in I \) be a critical point.
  (i) If \( f'(x) \) changes sign from positive to negative as \( x \) increases through \( c \), then \( c \) is called a point of local maxima.
  (ii) If \( f'(x) \) changes sign from negative to positive as \( x \) increases through \( c \), then \( c \) is a point of local minima.
  (iii) If \( f'(x) \) does not change sign as \( x \) increases through \( c \), then \( c \) is neither a point of local maxima nor a point of local minima. Such a point is called a point of inflexion.
• Second Derivative Test: Let \( f \) be a function defined on an interval \( I \) and let \( c \in I \).
  (i) \( x = c \) is a point of local maxima if \( f'(c) = 0 \) and the second derivative \( f''(c) < 0 \). In this case, \( f(c) \) is the local maximum value of \( f \).
  (ii) \( x = c \) is a point of local minima if \( f'(c) = 0 \) and the second derivative \( f''(c) > 0 \). In this case, \( f(c) \) is the local minimum value of \( f \).
  (iii) The test fails if both \( f'(c) = 0 \) and \( f''(c) = 0 \).

Very Short Answer Type Questions (1 Mark)

Question 1. The side of a square is increasing at the rate of 0.2 cm/sec. Find the rate of increase of perimeter of the square.
Answer:
Let the side of the square be \( x \).
We are given that the rate of increase of the side is:
\( \frac{dx}{dt} = 0.2 \text{ cm/sec} \).
The perimeter \( P \) of a square is given by the formula:
\( P = 4x \).
Differentiating both sides with respect to time \( t \):
\( \frac{dP}{dt} = \frac{d}{dt}(4x) = 4 \frac{dx}{dt} \).
Substituting the given value of \( \frac{dx}{dt} \):
\( \frac{dP}{dt} = 4(0.2) = 0.8 \text{ cm/sec} \).
Therefore, the perimeter is increasing at a rate of \( 0.8 \text{ cm/sec} \).
In simple words: The perimeter of a square is always 4 times its side length. Since the side increases by 0.2 cm every second, the perimeter must increase 4 times faster, which is 0.8 cm every second.

Exam Tip: Always state the units (like cm/sec or cm\(^2\)/sec) clearly in your final answer to ensure you do not lose minor presentation marks.

Question 2. The radius of the circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference?
Answer:
Let \( r \) be the radius of the circle.
We are given that the rate of increase of the radius is:
\( \frac{dr}{dt} = 0.7 \text{ cm/sec} \xb \).
The circumference \( C \) of a circle is given by:
\( C = 2\pi r \).
Differentiating both sides with respect to time \( t \):
\( \frac{dC}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \frac{dr}{dt} \).
Substituting the given value of \( \frac{dr}{dt} \) and using \( \pi \approx \frac{22}{7} \):
\( \frac{dC}{dt} = 2 \times \frac{22}{7} \times 0.7 = 2 \times 22 \times 0.1 = 4.4 \text{ cm/sec} \).
Therefore, the rate of increase of the circumference is \( 4.4 \text{ cm/sec} \).
In simple words: The circumference is \( 2\pi \) times the radius. Multiply the rate of the radius by \( 2\pi \) to get the rate of the circumference, which simplifies to 4.4 cm/sec.

Exam Tip: Substituting \( \pi = \frac{22}{7} \) helps simplify decimal rate problems when the rate is a multiple of 0.7 or 7.

Question 3. If the radius of a soap bubble is increasing at the rate of \( \frac{1}{2} \) cm/sec. At what rate its volume is increasing when the radius is 1 cm.
Answer:
Let the radius of the spherical soap bubble be \( r \).
We are given that the rate of increase of the radius is:
\( \frac{dr}{dt} = \frac{1}{2} \text{ cm/sec} \).
The volume \( V \) of a sphere is given by the formula:
\( V = \frac{4}{3}\pi r^3 \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( \frac{dV}{dt} = \frac{4}{3}\pi \left(3r^2 \frac{dr}{dt}\right) = 4\pi r^2 \frac{dr}{dt} \).
Substitute \( r = 1 \text{ cm} \) and \( \frac{dr}{dt} = \frac{1}{2} \text{ cm/sec} \):
\( \frac{dV}{dt} = 4\pi (1)^2 \left(\frac{1}{2}\right) = 2\pi \text{ cm}^3\text{/sec} \).
Therefore, the volume is increasing at the rate of \( 2\pi \text{ cm}^3\text{/sec} \).
In simple words: Differentiate the volume of a sphere to find its rate of change, then plug in the radius of 1 cm and the rate of 0.5 cm/sec to find the final volumetric rate of \( 2\pi \) cubic centimeters per second.

Exam Tip: Volumetric rates of change must always have cubic units per unit of time, such as cm\(^3\)/sec or m\(^3\)/sec.

Question 4. A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?
Answer:
Let \( r \) be the radius of the circular wave.
The speed of the wave is the rate of change of the radius:
\( \frac{dr}{dt} = 4 \text{ cm/sec} \).
The enclosed area \( A \) of the circle is:
\( A = \pi r^2 \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).
Substituting \( r = 10 \text{ cm} \) and \( \frac{dr}{dt} = 4 \text{ cm/sec} \):
\( \frac{dA}{dt} = 2\pi(10)(4) = 80\pi \text{ cm}^2\text{/sec} \).
Therefore, the enclosed area is increasing at a rate of \( 80\pi \text{ cm}^2\text{/sec} \).
In simple words: Differentiate the circle area equation to find the rate, then substitute the radius of 10 cm and the speed of 4 cm/sec to get the final area rate of \( 80\pi \) square centimeters per second.

Exam Tip: "Waves move in circles at a speed of..." always represents the rate of change of the radius, \( \frac{dr}{dt} \).

Very Short Answer Type Questions (1 Mark)

Question 5. The total revenue in rupees received from the sale of x units of a product is given by \( R(x) = 13x^2 + 26x + 15 \). Find the marginal revenue when x = 7.
Answer:
Marginal Revenue (\( MR \)) is defined as the rate of change of total revenue with respect to the number of units sold:
\( MR = \frac{dR}{dx} = \frac{d}{dx}(13x^2 + 26x + 15) = 26x + 26 \).
Now evaluate the marginal revenue at \( x = 7 \):
\( MR = 26(7) + 26 = 182 + 26 = 208 \).
Therefore, the marginal revenue when \( x = 7 \) is Rs. 208.
In simple words: Differentiate the revenue equation to get the marginal revenue function, and then substitute \( x = 7 \) to find the additional revenue earned from selling one more unit.

Exam Tip: "Marginal revenue" is always the first-order derivative of the total revenue function \( R(x) \).

Question 6. Find the maximum and minimum values of function f(x) = sin 2x + 5.
Answer:
We know that the sine function oscillates within a bounded range:
\( -1 \le \sin 2x \le 1 \).
Adding 5 to all parts of the inequality:
\( -1 + 5 \le \sin 2x + 5 \le 1 + 5 \)
\( 4 \le f(x) \le 6 \).
Therefore, the minimum value of \( f(x) \) is 4 and the maximum value of \( f(x) \) is 6.
In simple words: Since the lowest value of a sine wave is -1 and the highest is 1, adding 5 means the graph will oscillate between a low of 4 and a high of 6.

Exam Tip: For simple trigonometric functions, using the bounded range inequalities is much faster and less error-prone than taking derivative tests.

Question 7. Find the maximum and minimum values (if any) of the function f(x) = – |x – 1| + 7 \( \forall x \in R \).
Answer:
We know that the absolute value of any real number is non-negative:
\( |x - 1| \ge 0 \).
Multiplying both sides by -1 changes the inequality direction:
\( -|x - 1| \le 0 \).
Adding 7 to both sides:
\( -|x - 1| + 7 \le 7 \implies f(x) \le 7 \).
Thus, the maximum value of \( f(x) \) is 7 (which occurs when \( x = 1 \)).
Since \( -|x - 1| \) decreases without bound as \( x \to \pm\infty \), there is no minimum value.
Therefore, the maximum value is 7 and the minimum value does not exist.
In simple words: The absolute value term is always positive or zero. Because of the negative sign in front, the function's value can only go down from its peak of 7. It has no bottom limit, so there is no minimum.

Exam Tip: Absolute value graphs of the form \( -|x-a| + b \) always open downwards, meaning they have a global maximum at \( b \) and no minimum.

Question 8. Find the value of a for which the function f(x) = \( x^2 \) – 2ax + 6, x > 0 is strictly increasing.
Answer:
Differentiating the function \( f(x) \) with respect to \( x \):
\( f'(x) = 2x - 2a \).
For \( f(x) \) to be strictly increasing, we must have \( f'(x) > 0 \) for the given domain \( x > 0 \):
\( 2x - 2a > 0 \implies x > a \) for all \( x > 0 \).
For the condition \( x > a \) to hold true for all positive values of \( x \) (no matter how small), the constant \( a \) must be less than or equal to 0:
\( a \le 0 \).
Therefore, the value of \( a \) must satisfy \( a \le 0 \).
In simple words: Find the derivative, which represents the slope. For the slope to stay positive for all \( x > 0 \), the turning point of the parabola must lie at or to the left of \( x = 0 \), which means \( a \le 0 \).

Exam Tip: Be sure to write the strictly increasing condition as \( f'(x) > 0 \) and analyze it with respect to the given domain boundaries.

Question 9. Write the interval for which the function f(x) = cos x, \( 0 \le x \le 2\pi \) is decreasing.
Answer:
Differentiating the function \( f(x) \):
\( f'(x) = -\sin x \xb \).
For the function to be decreasing, we must have \( f'(x) \le 0 \):
\( -\sin x \le 0 \implies \sin x \ge 0 \).
In the given interval \( [0, 2\pi] \), the sine function is non-negative in the first and second quadrants:
\( x \in [0, \pi] \).
Therefore, the interval where \( f(x) \) is decreasing is \( [0, \pi] \).
In simple words: The derivative of \( \cos x \) is \( -\sin x \). For the slope to be negative (decreasing), \( \sin x \) must be positive, which happens in the interval from 0 to \( \pi \).

Exam Tip: Remember the quadrant sign behaviors of basic trigonometric functions; sine is positive in the interval \( [0, \pi] \) and negative in \( [\pi, 2\pi] \).

Question 10. What is the interval on which the function f(x) = \( \frac{\log x}{x} \), \( x \in (0, \infty) \) is increasing?
Answer:
Differentiating \( f(x) \) using the quotient rule:
\( f'(x) = \frac{\left(\frac{1}{x}\right) \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2} \).
For \( f(x) \) to be increasing, we must have \( f'(x) \ge 0 \):
\( \frac{1 - \log x}{x^2} \ge 0 \).
Since \( x^2 > 0 \) for all \( x \in (0, \infty) \), this inequality simplifies to:
\( 1 - \log x \ge 0 \implies \log x \le 1 \implies x \le e \).
Combining this with the given domain \( (0, \infty) \), we find:
\( x \in (0, e] \).
Therefore, the interval of increase is \( (0, e] \).
In simple words: Find the derivative using the quotient rule. Set the numerator to be greater than or equal to zero to ensure a positive slope, which shows the function increases for \( x \) values up to \( e \).

Exam Tip: Always intersect your solved inequality with the given domain of the logarithmic function, which restricts the values to positive real numbers.

Question 11. For which values of x, the functions \( y = x^4 - \frac{4}{3}x^3 \) is increasing?
Answer:
Differentiating \( y \) with respect to \( x \):
\( \frac{dy}{dx} = 4x^3 - 4x^2 = 4x^2(x - 1) \).
For the function to be increasing, we must have \( \frac{dy}{dx} \ge 0 \):
\( 4x^2(x - 1) \ge 0 \).
Since \( 4x^2 \ge 0 \) for all real numbers \( x \):
- For \( x \ne 0 \), we must have \( x - 1 \ge 0 \implies x \ge 1 \).
- For \( x = 0 \), the derivative is \( 0 \), which is also valid for non-strict increasing conditions.
Therefore, the function is increasing for all values \( x \ge 1 \).
In simple words: Find the derivative, factor it, and set it to be positive or zero. Since the squared term is always positive, the slope depends solely on \( x-1 \), making the function increase when \( x \ge 1 \).

Exam Tip: Factoring out common powers (like \( 4x^2 \)) helps isolate the term that actually determines the sign of the derivative.

Question 12. Write the interval for which the function f(x) = \( \frac{1}{x} \) is strictly decreasing.
Answer:
Differentiating \( f(x) \) with respect to \( x \):
\( f'(x) = -\frac{1}{x^2} \).
For the function to be strictly decreasing, we must have \( f'(x) < 0 \):
\( -\frac{1}{x^2} < 0 \).
Since \( x^2 > 0 \) for all non-zero real numbers, \( -\frac{1}{x^2} \) is strictly negative for all \( x \ne 0 \).
Therefore, the interval where \( f(x) \) is strictly decreasing is \( (-\infty, 0) \cup (0, \infty) \).
In simple words: The derivative of \( 1/x \) is always negative because the denominator is squared. This means the graph is constantly sloping downwards everywhere except at \( x = 0 \) where it is undefined.

Exam Tip: Do not write \( \mathbb{R} \) as the answer; \( x = 0 \) is not in the domain of the function, so it must be excluded from the interval.

Question 13. Find the sub-interval of the interval (0, \( \pi/2 \)) in which the function f(x) = sin 3x is increasing.
Answer:
Differentiating \( f(x) \):
\( f'(x) = 3 \cos 3x \).
For \( f(x) \) to be increasing, we must have \( f'(x) \ge 0 \):
\( 3 \cos 3x \ge 0 \implies \cos 3x \ge 0 \).
Given \( x \in (0, \pi/2) \), the angle \( 3x \in (0, 3\pi/2) \).
In this range, \( \cos 3x \) is positive in the first quadrant, i.e.:
\( 3x \in (0, \pi/2] \)

\( \implies x \in \left(0, \frac{\pi}{6}\right] \).
Therefore, the sub-interval of increase is \( \left(0, \frac{\pi}{6}\right] \).
In simple words: The derivative is \( 3\cos 3x \). For this to be positive, the angle \( 3x \) must lie in the first quadrant (up to \( \pi/2 \)), which means \( x \) must lie in the interval up to \( \pi/6 \).

Exam Tip: Scale your interval domain when working with multiple angles like \( 3x \) to correctly identify where trig signs change.

Question 14. Without using derivatives, find the maximum and minimum value of y = |3 sin x + 1|.
Answer:
We know the bounds of the sine function:
\( -1 \le \sin x \le 1 \).
Multiplying by 3:
\( -3 \le 3 \sin x \le 3 \).
Adding 1:
\( -2 \le 3 \sin x + 1 \le 4 \).
Taking the absolute value:
- The minimum value of an absolute value term is 0 (occurring when \( 3 \sin x + 1 = 0 \implies \sin x = -1/3 \), which is a valid value).
- The maximum value of the absolute value is \( \max(|-2|, |4|) = 4 \).
Therefore, the minimum value is 0 and the maximum value is 4.
In simple words: The expression inside the modulus ranges from -2 to 4. Since absolute value strips away negative signs, the lowest value is 0 and the highest value is 4.

Exam Tip: For absolute values of bounded functions, the minimum is 0 if the range contains both negative and positive numbers.

Question 15. If f(x) = ax + cos x is strictly increasing on R, find a.
Answer:
Differentiating the function \( f(x) \):
\( f'(x) = a - \sin x \).
For \( f(x) \) to be strictly increasing, we must have \( f'(x) > 0 \) for all \( x \in \mathbb{R} \):
\( a - \sin x > 0 \implies a > \sin x \).
Since the maximum value of \( \sin x \) is 1, this condition holds true for all real values of \( x \) if:
\( a > 1 \).
Therefore, \( a \) must be strictly greater than 1.
In simple words: The derivative is \( a - \sin x \). For this to stay positive, the constant \( a \) must be larger than the highest value of \( \sin x \) (which is 1), so \( a > 1 \).

Exam Tip: Set up your inequality with respect to the extreme values of the trigonometric terms to find the bounding range for \( a \).

Question 16. Write the interval in which the function f(x) = \( x^9 + 3x^7 \) + 64 is increasing.
Answer:
Differentiating \( f(x) \) with respect to \( x \):
\( f'(x) = 9x^8 + 21x^6 = 3x^6(3x^2 + 7) \).
We analyze the signs of the factors:
- \( 3x^6 \ge 0 \) for all real numbers \( x \).
- \( 3x^2 + 7 > 0 \) for all real numbers \( x \) (since \( 3x^2 \ge 0 \)).
Thus, \( f'(x) \ge 0 \) for all \( x \in \mathbb{R} \).
Therefore, the function is increasing on the entire set of real numbers \( \mathbb{R} \) (or \( (-\infty, \infty) \)).
In simple words: The derivative is \( 3x^6(3x^2 + 7) \). Since both factored parts contain even powers of \( x \) added to positive constants, the slope is never negative, making the function increase everywhere.

Exam Tip: Functions whose derivatives are always non-negative on \( \mathbb{R} \) are called monotonic increasing functions.

Question 17. What is the slope of the tangent to the curve f = \( x^3 \) – 5x + 3 at the point whose x co-ordinate is 2?
Answer:
The slope \( m \) of the tangent is given by the derivative of the function evaluated at the point:
\( m = f'(x) = \frac{d}{dx}(x^3 - 5x + 3) = 3x^2 - 5 \).
Now substitute \( x = 2 \):
\( m = 3(2)^2 - 5 = 12 - 5 = 7 \xb \).
Therefore, the slope of the tangent is 7.
In simple words: Find the derivative of the curve's equation to get the slope formula, and then substitute \( x = 2 \) to find the slope of the tangent line at that point.

Exam Tip: Finding the "slope of a tangent" is synonymous with evaluating the first-order derivative \( \frac{dy}{dx} \) at the given coordinate.

Question 18. At what point on the curve y = \( x^2 \) does the tangent make an angle of 45° with positive direction of the x-axis?
Answer:
The slope \( m \) of the tangent making an angle of \( 45^\circ \) is:
\( m = \tan 45^\circ = 1 \).
The slope of the tangent is also given by the derivative of the curve \( y = x^2 \):
\( m = \frac{dy}{dx} = 2x \).
Setting these equal:
\( 2x = 1 \implies x = \frac{1}{2} \).
Substitute \( x = \frac{1}{2} \) back into the curve equation \( y = x^2 \):
\( y = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
Therefore, the point on the curve is \( \left(\frac{1}{2}, \frac{1}{4}\right) \).
In simple words: An angle of \( 45^\circ \) means the slope is 1. Find where the derivative of \( x^2 \) equals 1, which gives the coordinate \( (1/2, 1/4) \).

Exam Tip: Always remember that the slope of a line is \( m = \tan\theta \), where \( \theta \) is the angle made with the positive \( x \)-axis.

Question 19. Find the point on the curve y = \( 3x^2 \) – 12x + 9 at which the tangent is parallel to x-axis.
Answer:
A tangent line is parallel to the \( x \)-axis if its slope is equal to zero.
Find the derivative of the curve:
\( \frac{dy}{dx} = 6x - 12 \).
Setting the slope to 0:
\( 6x - 12 = 0 \implies 6x = 12 \implies x = 2 \xb \).
Substitute \( x = 2 \) back into the curve equation to find \( y \):
\( y = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \).
Therefore, the point on the curve is \( (2, -3) \).
In simple words: "Parallel to the \( x \)-axis" means the slope is 0. Set the derivative to 0 to find \( x = 2 \), and plug this back into the original equation to get the point \( (2, -3) \).

Exam Tip: A flat horizontal tangent has slope 0, while a vertical tangent has an undefined slope (denominator of \( \frac{dy}{dx} \) is 0).

Question 20. What is the slope of the normal to the curve y = \( 5x^2 \) – 4 sin x at x = 0.
Answer:
First, find the derivative of the curve to get the slope of the tangent:
\( \frac{dy}{dx} = 10x - 4 \cos x \).
At \( x = 0 \), the slope of the tangent \( m_t \) is:
\( m_t = 10(0) - 4 \cos(0) = -4 \).
The slope of the normal \( m_n \) is perpendicular to the tangent:
\( m_n = -\frac{1}{m_t} = -\frac{1}{-4} = \frac{1}{4} \).
Therefore, the slope of the normal is \( \frac{1}{4} \).
In simple words: Find the slope of the tangent at \( x = 0 \) (which is -4). Since the normal is perpendicular to the tangent, take its negative reciprocal to find the normal slope of \( 1/4 \).

Exam Tip: Always use the perpendicular relation \( m_t \cdot m_n = -1 \) to convert tangent slopes into normal slopes.

Question 21. Find the point on the curve y = \( 3x^2 \) + 4 at which the tangent is perpendicular to the line with slope \( -\frac{1}{6} \).
Answer:
Let \( m_t \) be the slope of the tangent line.
Since the tangent is perpendicular to a line with slope \( -\frac{1}{6} \):
\( m_t \cdot \left(-\frac{1}{6}\right) = -1 \implies m_t = 6 \).
The slope of the tangent is also given by the derivative of the curve:
\( m_t = \frac{dy}{dx} = 6x \).
Setting these equal:
\( 6x = 6 \implies x = 1 \).
Substitute \( x = 1 \) back into the curve equation to find \( y \):
\( y = 3(1)^2 + 4 = 7 \).
Therefore, the point on the curve is \( (1, 7) \).
In simple words: Perpendicular to a slope of \( -1/6 \) means the tangent slope must be 6. Find where the derivative \( 6x \) equals 6, which gives \( x = 1 \), and plug it in to find \( (1, 7) \).

Exam Tip: Be sure to write the perpendicular condition \( m_1 \cdot m_2 = -1 \) clearly to justify how you solved for the target slope.

Question 22. Find the point on the curve y = \( x^2 \) where the slope of the tangent is equal to the y – co-ordinate.
Answer:
The slope \( m \) of the tangent is given by the derivative:
\( m = \frac{dy}{dx} = 2x \).
We are given that the slope is equal to the \( y \)-coordinate:
\( m = y \implies 2x = y \).
Substitute \( y = x^2 \) from the curve equation:
\( 2x = x^2 \)
\( \implies x^2 - 2x = 0 \)
\( \implies x(x - 2) = 0 \).
This gives:
- For \( x = 0 \): \( y = 0^2 = 0 \).
- For \( x = 2 \): \( y = 2^2 = 4 \).
Therefore, the points on the curve are \( (0, 0) \) and \( (2, 4) \).
In simple words: Set the derivative \( 2x \) equal to the \( y \) variable. Replace \( y \) with \( x^2 \) to solve the quadratic equation, which gives two coordinates: \( (0,0) \) and \( (2,4) \).

Exam Tip: Do not divide by \( x \) when solving \( x^2 = 2x \); doing so will lose the \( x = 0 \) root. Always factor the quadratic instead.

Question 23. If the curves y = \( 2e^x \) and y = \( ae^{-x} \) intersect orthogonally (cut at right angles), what is the value of a?
Answer:
Let \( (x_1, y_1) \) be the point of intersection.
At this point, the \( y \)-values must match:
\( y_1 = 2e^{x_1} = ae^{-x_1} \implies e^{2x_1} = \frac{a}{2} \).
For the first curve \( y = 2e^x \), the tangent slope is:
\( m_1 = \frac{dy}{dx} = 2e^x \implies \text{at intersection, } m_1 = 2e^{x_1} \).
For the second curve \( y = ae^{-x} \), the tangent slope is:
\( m_2 = \frac{dy}{dx} = -ae^{-x} \implies \text{at intersection, } m_2 = -ae^{-x_1} \).
Since they intersect orthogonally:
\( m_1 \cdot m_2 = -1 \)
\( (2e^{x_1}) \cdot \left(-ae^{-x_1}\right) = -1 \)
\( -2a = -1 \implies a = \frac{1}{2} \).
Therefore, the value of \( a \) is \( \frac{1}{2} \).
In simple words: Find the derivatives of both curves. Since they intersect at right angles, the product of their slopes is -1, which simplifies to give \( a = 1/2 \).

Exam Tip: Orthogonal intersection means that the product of their slopes at the point of intersection is \( -1 \).

Question 24. Find the slope of the normal to the curve y = \( 8x^2 \) – 3 at \( x = \frac{1}{4} \).
Answer:
First, find the derivative of the curve to get the slope of the tangent:
\( \frac{dy}{dx} = 16x \).
At \( x = \frac{1}{4} \), the tangent slope \( m_t \) is:
\( m_t = 16\left(\frac{1}{4}\right) = 4 \).
The slope of the normal \( m_n \) is perpendicular to the tangent:
\( m_n = -\frac{1}{m_t} = -\frac{1}{4} \).
Therefore, the slope of the normal is \( -\frac{1}{4} \).
In simple words: Find the slope of the tangent at \( x = 1/4 \) (which is 4). Taking its negative reciprocal gives the slope of the normal as \( -1/4 \).

Exam Tip: Be sure to write out the first derivative step explicitly before substituting the fractional point.

Question 25. Find the rate of change of the total surface area of a cylinder of radius r and height h with respect to radius when height is equal to the radius of the base of cylinder.
Answer:
The total surface area \( S \) of a cylinder is given by:
\( S = 2\pi r^2 + 2\pi rh \).
Differentiating \( S \) with respect to the radius \( r \) (treating \( h \) as a constant parameter):
\( \frac{dS}{dr} = \frac{d}{dr}(2\pi r^2 + 2\pi rh) = 4\pi r + 2\pi h \).
When the height is equal to the radius of the base (\( h = r \)):
\( \frac{dS}{dr} = 4\pi r + 2\pi r = 6\pi r \).
Therefore, the rate of change of the total surface area with respect to the radius is \( 6\pi r \).
In simple words: Differentiate the surface area equation with respect to \( r \), and then replace \( h \) with \( r \) to find the rate of change as \( 6\pi r \).

Exam Tip: Differentiate first with respect to the variable before substituting any momentary parameter relations (like \( h = r \)).

Question 26. Find the rate of change of the area of a circle with respect to its radius. How fast is the area changing w.r.t. its radius when its radius is 3 cm?
Answer:
The area \( A \) of a circle is given by:
\( A = \pi r^2 \xb \).
The rate of change of the area with respect to the radius \( r \) is:
\( \frac{dA}{dr} = 2\pi r \).
When \( r = 3 \text{ cm} \):
\( \frac{dA}{dr} = 2\pi(3) = 6\pi \text{ cm}^2/\text{cm} \).
Therefore, the area is changing at a rate of \( 6\pi \text{ cm}^2/\text{cm} \).
In simple words: Differentiate \( \pi r^2 \) with respect to \( r \) to get \( 2\pi r \), and substitute \( r = 3 \) to find the rate of change of \( 6\pi \).

Exam Tip: "With respect to its radius" means we differentiate directly with respect to \( r \), not with respect to time \( t \).

Question 27. For the curve y = (2x + 1)3 find the rate of change of slope at x = 1.
Answer:
The slope \( m \) of the curve is given by its first derivative:
\( m = \frac{dy}{dx} = 3(2x + 1)^2 \cdot \frac{d}{dx}(2x + 1) = 6(2x + 1)^2 \).
The rate of change of the slope is given by differentiating \( m \) with respect to \( x \):
\( \frac{dm}{dx} = \frac{d^2 y}{dx^2} = 12(2x + 1) \cdot 2 = 24(2x + 1) \).
At \( x = 1 \), the rate of change of slope is:
\( \frac{dm}{dx} = 24(2(1) + 1) = 24(3) = 72 \).
Therefore, the rate of change of the slope is 72.
In simple words: Find the first derivative to get the slope equation, differentiate it again to find the rate of change of the slope, and then plug in \( x = 1 \).

Exam Tip: The "rate of change of slope" is simply the second-order derivative \( \frac{d^2 y}{dx^2} \) of the curve.

Question 28. Find the slope of the normal to the curve x = 1 – a sin  ; y = b cos2 at \( \theta = \frac{\pi}{2} \).
Answer:
We find the parametric derivatives with respect to \( \theta \):
\( \frac{dx}{d\theta} = -a \cos\theta \).
\( \frac{dy}{d\theta} = 2b \cos\theta (-\sin\theta) = -2b \cos\theta \sin\theta \).
The slope of the tangent \( m_t \) is:
\( m_t = \frac{dy/d\theta}{dx/d\theta} = \frac{-2b \cos\theta \sin\theta}{-a \cos\theta} = \frac{2b}{a} \sin\theta \).
At \( \theta = \frac{\pi}{2} \):
\( m_t = \frac{2b}{a} \sin\left(\frac{\pi}{2}\right) = \frac{2b}{a} \).
The slope of the normal \( m_n \) is:
\( m_n = -\frac{1}{m_t} = -\frac{a}{2b} \).
Therefore, the slope of the normal is \( -\frac{a}{2b} \).
In simple words: Find the derivatives with respect to \( \theta \), divide them to find the tangent slope \( m_t \), evaluate at \( \theta = \pi/2 \), and take its negative reciprocal.

Exam Tip: Be sure to keep the negative sign during the reciprocal normal slope conversion step.

Question 29. If a manufacturer’s total cost function is C(x) = 1000 + 40x + \( x^2 \), where x is the out put, find the marginal cost for producing 20 units.
Answer:
Marginal Cost (\( MC \)) is the rate of change of total cost with respect to output \( x \):
\( MC = \frac{dC}{dx} = \frac{d}{dx}(1000 + 40x + x^2) = 40 + 2x \).
When output \( x = 20 \):
\( MC = 40 + 2(20) = 40 + 40 = 80 \).
Therefore, the marginal cost for producing 20 units is 80.
In simple words: Differentiate the cost equation to get the marginal cost formula, and substitute \( x = 20 \) to find the cost of producing one more unit.

Exam Tip: "Marginal cost" is always the first derivative of the cost function \( C(x) \).

Question 30. Find ‘a’ for which f (x) = a (x + sin x) is strictly increasing on R.
Answer:
Differentiating the function \( f(x) \) with respect to \( x \):
\( f'(x) = a(1 + \cos x) \).
For the function to be strictly increasing, we must have \( f'(x) > 0 \) for all \( x \in \mathbb{R} \):
\( a(1 + \cos x) > 0 \).
Since \( -1 \le \cos x \le 1 \), we have \( 1 + \cos x \ge 0 \) for all real numbers \( x \).
- For \( x \ne (2n+1)\pi \), we have \( 1 + \cos x > 0 \), which requires \( a > 0 \).
- At \( x = (2n+1)\pi \), \( f'(x) = 0 \).
Since the derivative is zero only at discrete isolated points, the function remains strictly increasing if:
\( a > 0 \).
Therefore, the function is strictly increasing on \( \mathbb{R} \) if \( a > 0 \).
In simple words: The derivative is \( a(1+\cos x) \). Since \( 1+\cos x \) is always positive or zero, the slope stays positive as long as \( a \) is strictly positive.

Exam Tip: A function can still be strictly increasing even if the derivative becomes zero at discrete isolated points, as long as it doesn't stay zero over an interval.

Short Answer Type Questions (4 Marks)

Question 31. A particle moves along the curve 6y = \( x^3 \) + 2. Find the points on the curve at which the y co-ordinate is changing 8 times as fast as the x co-ordinate.
Answer:
We are given the curve equation:
\( 6y = x^3 + 2 \).
Differentiating both sides with respect to time \( t \):
\( 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \).
We are given that the \( y \)-coordinate is changing 8 times as fast as the \( x \)-coordinate:
\( \frac{dy}{dt} = 8 \frac{dx}{dt} \).
Substitute this relation into the differentiated equation:
\( 6 \left(8 \frac{dx}{dt}\right) = 3x^2 \frac{dx}{dt} \)
\( 48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt} \).
Assuming \( \frac{dx}{dt} \ne 0 \), we divide both sides:
\( 3x^2 = 48 \)
\( \implies x^2 = 16 \)
\( \implies x = \pm 4 \).
- For \( x = 4 \):
\( 6y = 4^3 + 2 = 64 + 2 = 66 \)
\( \implies y = 11 \).
- For \( x = -4 \):
\( 6y = (-4)^3 + 2 = -64 + 2 = -62 \)
\( \implies y = -\frac{31}{3} \).
Therefore, the required points on the curve are \( (4, 11) \) and \( \left(-4, -\frac{31}{3}\right) \).
In simple words: Find the derivative of both sides with respect to time. Substitute \( dy/dt = 8 dx/dt \) to solve for \( x \), then plug those \( x \)-values back into the curve equation to find the corresponding points.

Exam Tip: Always show both cases for \( x = \pm 4 \); missing the negative root is a common error that will result in loss of marks.

Question 32. A ladder 5 metres long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 2 cm/sec. How fast is its height on the wall decreasing when the foot of the ladder is 4 metres away from the wall?
Answer:
Let \( x \) be the distance of the ladder's foot from the wall and \( y \) be the height of the ladder on the wall.
By the Pythagorean theorem:
\( x^2 + y^2 = 5^2 = 25 \) --- (1)
Differentiating both sides with respect to time \( t \):
\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \)
\( \implies x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \) --- (2)
We are given that the foot is pulled away at a rate of:
\( \frac{dx}{dt} = 2 \text{ cm/sec} = 0.02 \text{ m/sec} \).
When the foot is \( x = 4 \text{ m} \) away, find \( y \) from equation (1):
\( 4^2 + y^2 = 25 \)
\( \implies y^2 = 25 - 16 = 9 \)
\( \implies y = 3 \text{ m} \).
Now substitute these values into equation (2):
\( 4(2) + 3 \frac{dy}{dt} = 0 \) (using \( \text{cm/sec} \) units directly)
\( 3 \frac{dy}{dt} = -8 \)
\( \implies \frac{dy}{dt} = -\frac{8}{3} \text{ cm/sec} \xb \).
Therefore, the height on the wall is decreasing at the rate of \( \frac{8}{3} \text{ cm/sec} \).
In simple words: Write the relation using Pythagoras' theorem. Differentiate w.r.t. time, calculate the height when the base is 4 m, and solve for the rate of change of height, which is negative because it is decreasing.

Exam Tip: When stating the final "rate of decreasing", do not keep the negative sign in the text; write "decreasing at the rate of \( \frac{8}{3} \text{ cm/sec} \)".

Question 33. A balloon which always remain spherical is being inflated by pumping in 900 cubic cm of a gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Answer:
Let \( V \) be the volume and \( r \) be the radius of the spherical balloon.
The volume is given by:
\( V = \frac{4}{3}\pi r^3 \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \).
We are given that the rate of change of volume is:
\( \frac{dV}{dt} = 900 \text{ cm}^3/\text{sec} \).
Substitute \( r = 15 \text{ cm} \) and \( \frac{dV}{dt} = 900 \):
\( 900 = 4\pi (15)^2 \frac{dr}{dt} \)
\( 900 = 900\pi \frac{dr}{dt} \)
\( \implies \frac{dr}{dt} = \frac{1}{\pi} \text{ cm/sec} \).
Therefore, the radius is increasing at a rate of \( \frac{1}{\pi} \text{ cm/sec} \).
In simple words: Write out the volume formula for a sphere, differentiate it, plug in the given inflation rate and radius, and solve for the rate of increase of the radius.

Exam Tip: Be sure to keep \( \pi \) in the denominator of your final answer; there is no need to write it as a decimal value unless specified.

Question 34. A man 2 meters high walks at a uniform speed of 5 km/hr away from a lamp post 6 metres high. Find the rate at which the length of his shadow increases.
Answer:
Let \( x \) be the distance of the man from the lamp post, and \( s \) be the length of his shadow.
By similar triangles:
\( \frac{s}{2} = \frac{x + s}{6} \)
\( 6s = 2x + 2s \)
\( \implies 4s = 2x \)
\( \implies s = \frac{x}{2} \).
Differentiating both sides with respect to time \( t \):
\( \frac{ds}{dt} = \frac{1}{2} \frac{dx}{dt} \).
Given that the speed of the man is \( \frac{dx}{dt} = 5 \text{ km/hr} \):
\( \frac{ds}{dt} = \frac{1}{2}(5) = 2.5 \text{ km/hr} \).
Therefore, the rate at which the length of his shadow increases is \( 2.5 \text{ km/hr} \).
In simple words: Use similar triangles to establish a relationship between the shadow length \( s \) and the distance walked \( x \). The derivative shows that the shadow grows at exactly half the speed of the walking man.

Exam Tip: Drawing a quick diagram illustrating similar triangles is helpful to make your steps visually clear to the examiner.

Question 35. Water is running out of a conical funnel at the rate of 5 \( \text{cm}^3 \)/sec. If the radius of the base of the funnel is 10 cm and altitude is 20 cm, find the rate at which the water level is dropping when it is 5 cm from the top.
Answer:
Let \( r \) be the radius and \( h \) be the height of the water level from the vertex of the cone.
The semi-vertical angle is \( \alpha \), where:
\( \tan\alpha = \frac{\text{Radius}}{\text{Altitude}} = \frac{10}{20} = \frac{1}{2} \).
For any water level height \( h \), the radius is:
\( r = h \tan\alpha = \frac{h}{2} \).
The volume \( V \) of water in the conical funnel is:
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12} \).
Differentiating both sides with respect to time \( t \):
\( \frac{dV}{dt} = \frac{3\pi h^2}{12} \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt} \).
We are given that the rate at which water is running out is \( \frac{dV}{dt} = -5 \text{ cm}^3/\text{sec} \).
When the water level is 5 cm from the top, its height from the vertex is:
\( h = 20 - 5 = 15 \text{ cm} \).
Substitute these values into the differentiated equation:
\( -5 = \frac{\pi (15)^2}{4} \frac{dh}{dt} \)
\( -20 = 225\pi \frac{dh}{dt} \)
\( \implies \frac{dh}{dt} = -\frac{20}{225\pi} = -\frac{4}{45\pi} \text{ cm/sec} \).
Therefore, the water level is dropping at a rate of \( \frac{4}{45\pi} \text{ cm/sec} \).
In simple words: Relate the radius and height of the cone using trigonometry, rewrite the volume in terms of height, differentiate, and solve for the rate when the height is 15 cm.

Exam Tip: Be careful with the height description. "5 cm from the top" means the actual depth of water from the vertex is \( 20 - 5 = 15 \text{ cm} \).

Question 36. The length x of a rectangle is decreasing at the rate of 5 cm/sec and the width y is increasing as the rate of 4 cm/sec when x = 8 cm and y = 6 cm. Find the rate of change of (a) Perimeter (b) Area of the rectangle.
Answer:
We are given that:
\( \frac{dx}{dt} = -5 \text{ cm/sec} \) (decreasing) and \( \frac{dy}{dt} = 4 \text{ cm/sec} \) (increasing).
We evaluate the rates when \( x = 8 \text{ cm} \) and \( y = 6 \text{ cm} \):

(a) Perimeter \( P = 2(x + y) \).
Differentiating both sides with respect to \( t \):
\( \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2 \text{ cm/sec} \).
Thus, the perimeter is decreasing at the rate of \( 2 \text{ cm/sec} \).

(b) Area \( A = xy \).
Differentiating both sides with respect to \( t \) using the product rule:
\( \frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt} = 8(4) + 6(-5) = 32 - 30 = 2 \text{ cm}^2/\text{sec} \xb \).
Thus, the area is increasing at the rate of \( 2 \text{ cm}^2/\text{sec} \).
In simple words: Use the perimeter and area formulas, differentiate them using calculus rules, substitute the rates (remembering a negative rate for decreasing), and find the final answers.

Exam Tip: Write down the product rule formula clearly for part (b) before plugging in any coordinates; this avoids simple arithmetic mistakes.

Question 37. Sand is pouring from a pipe at the rate of 12c.c/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when height is 4 cm?
Answer:
Let \( r \) be the radius and \( h \) be the height of the sand cone.
The volume \( V \) of a cone is:
\( V = \frac{1}{3}\pi r^2 h \).
We are given that the height is always one-sixth of the radius:
\( h = \frac{r}{6} \implies r = 6h \).
Substitute \( r \) back into the volume formula to express \( V \) in terms of \( h \) only:
\( V = \frac{1}{3}\pi (6h)^2 h = 12\pi h^3 \).
Differentiating both sides with respect to time \( t \):
\( \frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt} \).
Given that the rate of pouring is \( \frac{dV}{dt} = 12 \text{ cm}^3/\text{sec} \):
When \( h = 4 \text{ cm} \):
\( 12 = 36\pi (4)^2 \frac{dh}{dt} \)
\( 12 = 576\pi \frac{dh}{dt} \)
\( \implies \frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi} \text{ cm/sec} \).
Therefore, the height is increasing at a rate of \( \frac{1}{48\pi} \text{ cm/sec} \).
In simple words: Replace the radius variable \( r \) with \( 6h \) in the cone volume formula, differentiate, plug in the given pouring rate and height, and solve for the rate of increase of height.

Exam Tip: Always convert your equations into a single variable first (height \( h \) in this case) before differentiating to keep the calculus simple.

Question 38. The area of an expanding rectangle is increasing at the rate of 48 \( \text{cm}^2 \)/ sec. The length of the rectangle is always equal to the square of the breadth. At what rate is the length increasing at the instant when the breadth is 4.5 cm?
Answer:
Let \( x \) be the breadth and \( y \) be the length of the rectangle.
We are given that \( y = x^2 \).
The area \( A \) of the rectangle is:
\( A = xy = x(x^2) = x^3 \).
Differentiating \( A \) with respect to \( t \):
\( \frac{dA}{dt} = 3x^2 \frac{dx}{dt} \).
We are given that \( \frac{dA}{dt} = 48 \text{ cm}^2/\text{sec} \).
At the instant when breadth \( x = 4.5 \text{ cm} = \frac{9}{2} \text{ cm} \):
\( 48 = 3\left(\frac{9}{2}\right)^2 \frac{dx}{dt} \)
\( 48 = \frac{243}{4} \frac{dx}{dt} \)
\( \implies \frac{dx}{dt} = \frac{192}{243} = \frac{64}{81} \text{ cm/sec} \).
Now, find the rate of change of length \( y = x^2 \) using the chain rule:
\( \frac{dy}{dt} = 2x \frac{dx}{dt} \).
At the instant when \( x = \frac{9}{2} \) and \( \frac{dx}{dt} = \frac{64}{81} \):
\( \frac{dy}{dt} = 2\left(\frac{9}{2}\right) \left(\frac{64}{81}\right) = 9 \left(\frac{64}{81}\right) = \frac{64}{9} \text{ cm/sec} \).
Therefore, the rate of increase of length is \( \frac{64}{9} \text{ cm/sec} \) (or approximately \( 7.11 \text{ cm/sec} \)).
In simple words: Express the area in terms of breadth \( x \), find \( dx/dt \) using the given area rate, and then differentiate the length equation \( y = x^2 \) to solve for \( dy/dt \).

Exam Tip: Working with fractions like \( 9/2 \) instead of decimals prevents rounding mistakes and keeps your algebraic steps clean.

Question 39. Find a point on the curve y = (x – 3)2 where the tangent is parallel to the line joining the points (4, 1) and (3, 0).
Answer:
The slope \( m \) of the line joining the points \( (4, 1) \) and \( (3, 0) \) is:
\( m = \frac{1 - 0}{4 - 3} = 1 \).
The slope of the tangent to the curve \( y = (x - 3)^2 \) is given by its derivative:
\( \frac{dy}{dx} = 2(x - 3) \).
Since the tangent is parallel to the line, their slopes must be equal:
\( 2(x - 3) = 1 \)
\( x - 3 = \frac{1}{2} \implies x = \frac{7}{2} \).
Substitute \( x = \frac{7}{2} \) back into the curve equation to find \( y \):
\( y = \left(\frac{7}{2} - 3\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
Therefore, the required point on the curve is \( \left(\frac{7}{2}, \frac{1}{4}\right) \).
In simple words: Find the slope of the line (which is 1). Set the derivative of the curve's equation equal to 1 to find the \( x \)-coordinate, and plug it back in to find the point.

Exam Tip: Remember that parallel lines always have equal slopes, so \( \frac{dy}{dx} \) must equal the slope of the secant line.

Question 40. Find the equation of all lines having slope zero which are tangents to the curve y = \( \frac{1}{x^2 - 2x + 3} \).
Answer:
We are given that the slope of the tangent line is zero:
\( \frac{dy}{dx} = 0 \).
Differentiating the curve equation \( y = (x^2 - 2x + 3)^{-1} \):
\( \frac{dy}{dx} = -(x^2 - 2x + 3)^{-2} \cdot (2x - 2) = \frac{-(2x - 2)}{(x^2 - 2x + 3)^2} \).
Setting this derivative equal to 0:
\( \frac{-(2x - 2)}{(x^2 - 2x + 3)^2} = 0 \implies 2x - 2 = 0 \implies x = 1 \).
Substitute \( x = 1 \) back into the curve equation to find \( y \):
\( y = \frac{1}{1^2 - 2(1) + 3} = \frac{1}{2} \).
The equation of the tangent line with slope 0 passing through \( \left(1, \frac{1}{2}\right) \) is:
\( y - \frac{1}{2} = 0(x - 1) \implies y = \frac{1}{2} \).
Therefore, the equation of the required tangent line is \( y = \frac{1}{2} \).
In simple words: Set the derivative of the curve to 0 to find where the tangent is horizontal. This gives \( x = 1 \), which corresponds to \( y = 1/2 \), making the tangent line equation \( y = 1/2 \).

Exam Tip: A tangent with slope 0 is always a horizontal line of the form \( y = c \), where \( c \) is the \( y \)-coordinate of the point of contact.

Question 41. Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.
Answer:
Let \( (x_1, y_1) \) be the point of intersection of the two curves.
Since the point lies on both curves:
\( x_1 = y_1^2 \) and \( x_1 y_1 = k \implies (y_1^2) y_1 = k \implies y_1^3 = k \implies y_1 = k^{1/3} \).
Also, \( x_1 = y_1^2 = k^{2/3} \).
Differentiate \( x = y^2 \) with respect to \( x \):
\( 1 = 2y y_1' \implies y_1' = \frac{1}{2y_1} \).
Differentiate \( xy = k \) with respect to \( x \):
\( y + x y_2' = 0 \implies y_2' = -\frac{y_1}{x_1} \).
For the curves to cut at right angles (orthogonally), the product of their slopes must be -1:
\( y_1' \cdot y_2' = -1 \)
\( \left(\frac{1}{2y_1}\right) \left(-\frac{y_1}{x_1}\right) = -1 \)
\( -\frac{1}{2x_1} = -1 \implies 2x_1 = 1 \implies x_1 = \frac{1}{2} \).
Substitute \( x_1 = k^{2/3} \):
\( k^{2/3} = \frac{1}{2} \).
Cubing both sides of the equation:
\( \left(k^{2/3}\right)^3 = \left(\frac{1}{2}\right)^3 \)
\( k^2 = \frac{1}{8} \implies 8k^2 = 1 \).
Hence proved.
In simple words: Find where the curves cross, differentiate both curves to find their slopes, multiply the slopes and set the product to -1 (perpendicular condition), and solve to find \( 8k^2 = 1 \).

Exam Tip: Be sure to write the orthogonality criterion \( m_1 \cdot m_2 = -1 \) clearly to justify your steps to the examiner.

Question 42. Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.
Answer:
We differentiate the curve equation \( a y^2 = x^3 \) with respect to \( x \):
\( 2ay \frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{2ay} \xb \).
At the given point \( (am^2, am^3) \), the slope of the tangent \( m_t \) is:
\( m_t = \frac{3(am^2)^2}{2a(am^3)} = \frac{3a^2 m^4}{2a^2 m^3} = \frac{3}{2}m \).
The slope of the normal \( m_n \) is perpendicular to the tangent:
\( m_n = -\frac{1}{m_t} = -\frac{2}{3m} \).
The equation of the normal line passing through \( (am^2, am^3) \) is:
\( y - am^3 = -\frac{2}{3m}(x - am^2) \)
\( 3m(y - am^3) = -2(x - am^2) \)
\( 3my - 3am^4 = -2x + 2am^2 \)
\( 2x + 3my - am^2(2 + 3m^2) = 0 \).
Therefore, the equation of the normal is \( 2x + 3my - am^2(2 + 3m^2) = 0 \).
In simple words: Differentiate the curve's equation to find the slope of the tangent, find the slope of the normal using negative reciprocals, and write the straight line equation using the point-slope formula.

Exam Tip: Group your constant terms cleanly (like factoring out \( am^2 \)) at the end to write the final equation in standard \( Ax + By + C = 0 \) form.

Question 43. Show that the curves 4x = y2 and 4xy = k cut as right angles if k2 = 512.
Answer:
Let \( (x_1, y_1) \) be the point of intersection of the two curves.
Since \( 4x_1 = y_1^2 \), substituting \( y_1 = \frac{k}{4x_1} \) from the second curve:
\( 4x_1 = \left(\frac{k}{4x_1}\right)^2 = \frac{k^2}{16x_1^2} \implies 64x_1^3 = k^2 \implies x_1^3 = \frac{k^2}{64} \).
Differentiate \( 4x = y^2 \) with respect to \( x \):
\( 4 = 2y y_1' \implies y_1' = \frac{2}{y_1} \xb \).
Differentiate \( 4xy = k \) with respect to \( x \):
\( 4y + 4x y_2' = 0 \implies y_2' = -\frac{y_1}{x_1} \).
For the curves to cut at right angles, the product of their slopes must be -1:
\( y_1' \cdot y_2' = -1 \)
\( \left(\frac{2}{y_1}\right) \left(-\frac{y_1}{x_1}\right) = -1 \)
\( -\frac{2}{x_1} = -1 \implies x_1 = 2 \).
Substitute \( x_1 = 2 \) into our intersection relation:
\( k^2 = 64(2^3) = 64 \times 8 = 512 \).
Hence proved.
In simple words: Find where the curves cross, differentiate both curves to find their slopes, multiply the slopes and set the product to -1, and solve to find \( k^2 = 512 \).

Exam Tip: This proof is identical in method to Question 41. Make sure you follow the same logical steps of finding derivatives and setting their product to -1.

Question 44. Find the equation of the tangent to the curve y = \( \sqrt{3x - 2} \) which is parallel to the line 4x – y + 5 = 0.
Answer:
The slope of the given line \( 4x - y + 5 = 0 \) is:
\( m = 4 \).
Since the tangent is parallel to this line, the slope of the tangent is also \( m_t = 4 \).
Differentiating the curve equation \( y = \sqrt{3x - 2} \):
\( \frac{dy}{dx} = \frac{1}{2\sqrt{3x - 2}} \cdot 3 = \frac{3}{2\sqrt{3x - 2}} \).
Setting the derivative equal to 4:
\( \frac{3}{2\sqrt{3x - 2}} = 4 \)
\( 8\sqrt{3x - 2} = 3 \implies \sqrt{3x - 2} = \frac{3}{8} \).
Squaring both sides of the equation:
\( 3x - 2 = \frac{9}{64} \implies 3x = 2 + \frac{9}{64} = \frac{137}{64} \implies x = \frac{137}{192} \).
Now find the \( y \)-coordinate using \( y = \sqrt{3x - 2} \):
\( y = \frac{3}{8} \).
The point of contact is \( \left(\frac{137}{192}, \frac{3}{8}\right) \).
The equation of the tangent line is:
\( y - \frac{3}{8} = 4\left(x - \frac{137}{192}\right) \)
\( y - \frac{3}{8} = 4x - \frac{137}{48} \).
Multiplying both sides by 48 to simplify:
\( 48y - 18 = 192x - 137 \)
\( 192x - 48y - 119 = 0 \).
Therefore, the equation of the tangent is \( 192x - 48y - 119 = 0 \).
In simple words: Parallel to the line means the tangent slope must be 4. Set the derivative of the curve to 4 to find the \( x \)-coordinate, get the corresponding \( y \), and write the line equation using the point-slope formula.

Exam Tip: Be careful with fraction simplification. Keep the terms as fractions rather than converting them to decimals, as standard board answers expect integer coefficients in the final line equation.

Question 45. Find the equation of the tangent to the curve \( \sqrt{x} \) + \( \sqrt{y} \) = \( \sqrt{a} \) at the point \( \left(\frac{a}{4}, \frac{a}{4}\right) \).
Answer:
Differentiating the given curve equation with respect to \( x \):
\( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \)
\( \frac{1}{\sqrt{y}} \frac{dy}{dx} = -\frac{1}{\sqrt{x}} \implies \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \).
At the given point \( \left(\frac{a}{4}, \frac{a}{4}\right) \), the slope of the tangent \( m_t \) is:
\( m_t = -\frac{\sqrt{a/4}}{\sqrt{a/4}} = -1 \).
The equation of the tangent line passing through \( \left(\frac{a}{4}, \frac{a}{4}\right) \) is:
\( y - \frac{a}{4} = -1\left(x - \frac{a}{4}\right) \)
\( y - \frac{a}{4} = -x + \frac{a}{4} \)
\( x + y = \frac{a}{2} \).
Therefore, the equation of the tangent is \( 2x + 2y - a = 0 \).
In simple words: Differentiate implicitly to find the slope formula, evaluate at the given point to find the slope of -1, and write the straight line equation using the point-slope formula.

Exam Tip: Watch out for the derivative of a constant term (like \( \sqrt{a} \)), which is always 0. A very common error is to write \( \frac{1}{2\sqrt{a}} \).

Question 46. Find the points on the curve 4y = x3 where slope of the tangent is 16/3.
Answer:
Given the curve equation:
\( 4y = x^3 \).
Differentiating both sides with respect to \( x \):
\( 4 \frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{4} \).
We are given that the slope of the tangent is \( \frac{16}{3} \):
\( \frac{3x^2}{4} = \frac{16}{3} \)
\( 9x^2 = 64 \implies x^2 = \frac{64}{9} \implies x = \pm \frac{8}{3} \).
- For \( x = \frac{8}{3} \):
\( 4y = \left(\frac{8}{3}\right)^3 = \frac{512}{27} \implies y = \frac{128}{27} \xb \).
- For \( x = -\frac{8}{3} \):
\( 4y = \left(-\frac{8}{3}\right)^3 = -\frac{512}{27} \implies y = -\frac{128}{27} \).
Therefore, the required points on the curve are \( \left(\frac{8}{3}, \frac{128}{27}\right) \) and \( \left(-\frac{8}{3}, -\frac{128}{27}\right) \).
In simple words: Differentiate the curve, set the derivative equal to \( 16/3 \) to solve for \( x \), and then plug both positive and negative \( x \)-values back in to find the coordinates.

Exam Tip: Don't forget to evaluate both the positive and negative roots when taking the square root of \( x^2 \).

Question 47. Show that x/a + y/b = 1 touches the curve y = be–x/a at the point where the curve crosses the y-axis.
Answer:
The curve crosses the \( y \)-axis where \( x = 0 \).
Substitute \( x = 0 \) into the curve equation to find the point of intersection:
\( y = b e^0 = b \).
So, the point of intersection is \( (0, b) \).
Now find the derivative of the curve \( y = b e^{-x/a} \) to find the tangent slope:
\( \frac{dy}{dx} = b e^{-x/a} \left(-\frac{1}{a}\right) = -\frac{b}{a} e^{-x/a} \).
At the point \( (0, b) \):
\( m_t = -\frac{b}{a} e^0 = -\frac{b}{a} \).
The equation of the tangent passing through \( (0, b) \) with slope \( -\frac{b}{a} \) is:
\( y - b = -\frac{b}{a} (x - 0) \)
\( a(y - b) = -bx \implies bx + ay = ab \).
Dividing both sides by the product \( ab \):
\( \frac{x}{a} + \frac{y}{b} = 1 \).
Since this is exactly equal to the given line, the line touches the curve.
Hence proved.
In simple words: Find where the curve crosses the \( y \)-axis (at \( x=0, y=b \)). Calculate the slope of the tangent at this point, write the tangent line equation, and show that it matches the given line.

Exam Tip: The phrase "touches the curve" is synonymous with "is tangent to the curve" at that specific point.

Question 48. Find the equation of the tangent to the curve given by x = a sin3t, y = b cos3 t at a point where t = \( \pi/2 \).
Answer:
First, find the point of contact when \( t = \frac{\pi}{2} \):
\( x = a \sin^3\left(\frac{\pi}{2}\right) = a(1)^3 = a \).
\( y = b \cos^3\left(\frac{\pi}{2}\right) = b(0)^3 = 0 \).
So, the point of contact is \( (a, 0) \).
Now find the parametric derivatives with respect to \( t \):
\( \frac{dx}{dt} = 3a \sin^2 t \cos t \).
\( \frac{dy}{dt} = 3b \cos^2 t (-\sin t) = -3b \cos^2 t \sin t \).
The slope of the tangent \( m_t \) is:
\( m_t = \frac{dy/dt}{dx/dt} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{b}{a} \cot t \).
At \( t = \frac{\pi}{2} \):
\( m_t = -\frac{b}{a} \cot\left(\frac{\pi}{2}\right) = 0 \).
The equation of the tangent line passing through \( (a, 0) \) with slope 0 is:
\( y - 0 = 0(x - a) \implies y = 0 \).
Therefore, the equation of the tangent is \( y = 0 \).
In simple words: Find the coordinates at \( t = \pi/2 \), calculate the parametric derivative to find the tangent slope of 0, and write the horizontal line equation.

Exam Tip: Be comfortable with trig values; \( \cot(\pi/2) \) is 0, which makes the tangent slope flat and horizontal.

Question 49. Find the intervals in which the function f(x) = log (1 + x) – \( \frac{x}{1 + x} \), x > –1 is increasing or decreasing.
Answer:
Differentiating the function \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{1}{1 + x} - \frac{1 \cdot (1 + x) - x \cdot 1}{(1 + x)^2} = \frac{1}{1 + x} - \frac{1}{(1 + x)^2} = \frac{1+x-1}{(1+x)^2} = \frac{x}{(1+x)^2} \).
Now analyze the signs of the derivative:
Since \( (1 + x)^2 > 0 \) for all \( x > -1 \), the sign of \( f'(x) \) depends solely on the numerator \( x \):
- For the function to be increasing, \( f'(x) \ge 0 \implies x \ge 0 \).
Thus, the function is increasing in \( [0, \infty) \).
- For the function to be decreasing, \( f'(x) \le 0 \implies x \le 0 \).
Taking the domain \( x > -1 \) into account, the function is decreasing in \( (-1, 0] \).
In simple words: Differentiate the function, simplify using a common denominator, and note that the sign depends only on \( x \). This tells us the function decreases when \( x \) is negative and increases when \( x \) is positive.

Exam Tip: Always remember to incorporate the given domain constraint \( x > -1 \) when writing your final decreasing interval.

Question 50. Find the intervals in which the function f(x) = x3 – 12x2 + 36x + 17 is (a) Increasing (b) Decreasing.
Answer:
Differentiating \( f(x) \) with respect to \( x \):
\( f'(x) = 3x^2 - 24x + 36 = 3(x^2 - 8x + 12) = 3(x - 2)(x - 6) \).
The critical points are \( x = 2 \) and \( x = 6 \).
- (a) Increasing: we must have \( f'(x) \ge 0 \implies (x - 2)(x - 6) \ge 0 \).
This holds true when \( x \in (-\infty, 2] \cup [6, \infty) \).
- (b) Decreasing: we must have \( f'(x) \le 0 \implies (x - 2)(x - 6) \le 0 \).
This holds true when \( x \in [2, 6] \).
In simple words: Find the derivative, factor the quadratic expression, and find the critical roots. Use the interval sign test to identify where the slope is positive (increasing) or negative (decreasing).

Exam Tip: Writing the intervals with closed brackets at critical points is standard practice unless the question specifies "strictly" increasing/decreasing.

Question 51. Prove that the function f(x) = x2 – x + 1 is neither increasing nor decreasing in [0, 1].
Answer:
Differentiating the function \( f(x) \) with respect to \( x \):
\( f'(x) = 2x - 1 \).
Set \( f'(x) = 0 \implies x = \frac{1}{2} \).
- For \( x \in [0, 1/2) \), the derivative is negative: \( f'(x) < 0 \), which means the function is decreasing.
- For \( x \in (1/2, 1] \), the derivative is positive: \( f'(x) > 0 \), which means the function is increasing.
Since the derivative changes sign within the interval \( [0, 1] \), the function is neither strictly increasing nor strictly decreasing across the entire interval \( [0, 1] \).
Hence proved.
In simple words: The derivative \( 2x-1 \) is negative in the first half of the interval and positive in the second half. Since the graph goes down and then up within \( [0,1] \), it cannot be described as purely increasing or decreasing overall.

Exam Tip: State clearly that the derivative changes sign within the interval; this is the key phrase examiners look for to verify your proof.

Question 52. Find the intervals on which the function f(x) = \( \frac{x}{x^2 + 1} \) is decreasing.
Answer:
Differentiating the function using the quotient rule:
\( f'(x) = \frac{1 \cdot (x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} = \frac{(1 - x)(1 + x)}{(x^2 + 1)^2} \).
For the function to be decreasing, we must have \( f'(x) \le 0 \):
\( \frac{(1 - x)(1 + x)}{(x^2 + 1)^2} \le 0 \).
Since the denominator \( (x^2 + 1)^2 \) is always strictly positive, this reduces to:
\( 1 - x^2 \le 0 \implies x^2 \ge 1 \implies x \in (-\infty, -1] \cup [1, \infty) \).
Therefore, the function is decreasing in the intervals \( (-\infty, -1] \cup [1, \infty) \xb \).
In simple words: Differentiate the function using the quotient rule. Since the bottom part is squared and always positive, the sign of the slope depends solely on \( 1-x^2 \). This is negative when \( x \) is outside the range of -1 and 1.

Exam Tip: Be sure to keep the denominator term squared throughout your quotient rule steps to maintain complete mathematical accuracy.

Question 53. Prove that \( f(x) = \frac{x^3}{3} - x^2 + 9x \), \( x \in [1, 2] \) is strictly increasing. Hence find the minimum value of f (x).
Answer:
Differentiating the function \( f(x) \):
\( f'(x) = x^2 - 2x + 9 = (x - 1)^2 + 8 \).
Since \( (x - 1)^2 \ge 0 \) for all real values of \( x \), we have:
\( f'(x) \ge 8 > 0 \) for all \( x \in \mathbb{R} \).
Since \( f'(x) \) is strictly positive on the interval \( [1, 2] \), the function \( f(x) \) is strictly increasing on \( [1, 2] \).
Because the function is strictly increasing, its lowest (minimum) value on the interval must occur at the left endpoint \( x = 1 \):
\( f(1) = \frac{1^3}{3} - (1)^2 + 9(1) = \frac{1}{3} - 1 + 9 = 8 + \frac{1}{3} = \frac{25}{3} \).
Therefore, the minimum value is \( \frac{25}{3} \).
In simple words: Differentiate and rewrite as a perfect square plus a constant, which proves the slope is always positive (increasing). Since the graph is constantly rising, the minimum value must occur at the very beginning of the interval (\( x=1 \)).

Exam Tip: Expressing a quadratic derivative as a completed square \( (x-h)^2 + k \) is a highly reliable way to prove that a derivative is always positive.

Question 54. Find the intervals in which the function f(x) = sin4x + cos4x, 0 \( \le \) x \( \le \) \( \frac{\pi}{2} \) is increasing or decreasing.
Answer:
We first simplify the function algebraically:
\( f(x) = \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x \).
Differentiating both sides with respect to \( x \):
\( f'(x) = -\frac{1}{2} \cdot 2\sin 2x \cos 2x \cdot 2 = -2\sin 2x \cos 2x = -\sin 4x \xb \).
Given the domain \( 0 \le x \le \pi/2 \), the multiple angle is bounded by \( 0 \le 4x \le 2\pi \).
- For the function to be decreasing:
\( f'(x) \le 0 \implies -\sin 4x \le 0 \implies \sin 4x \ge 0 \).
The sine function is positive in the first and second quadrants, i.e., \( 4x \in [0, \pi] \implies x \in [0, \pi/4] \).
- For the function to be increasing:
\( f'(x) \ge 0 \implies \sin 4x \le 0 \).
The sine function is negative in the third and fourth quadrants, i.e., \( 4x \in [\pi, 2\pi] \implies x \in [\pi/4, \pi/2] \).
Therefore, the function is decreasing in \( [0, \pi/4] \) and increasing in \( [\pi/4, \pi/2] \).
In simple words: Simplify the equation using trigonometric double-angle identities to get a derivative of \( -\sin 4x \). Test where this sine term is positive or negative to find the increasing and decreasing intervals.

Exam Tip: Transforming powers like \( \sin^4 x + \cos^4 x \) using double angles is a great way to avoid long and complicated chain rule steps during differentiation.

Question 55. Find the least value of 'a' such that the function f(x) = x2 + ax + 1 is strictly increasing on (1, 2).
Answer:
Differentiating the function with respect to \( x \):
\( f'(x) = 2x + a \xb \).
For \( f(x) \) to be strictly increasing on \( (1, 2) \), the derivative must be strictly positive:
\( 2x + a > 0 \implies a > -2x \) for all \( x \in (1, 2) \).
Since \( 1 < x < 2 \), the term \( -2x \) lies in the interval:
\( -4 < -2x < -2 \).
For \( a \) to be strictly greater than \( -2x \) for all values of \( x \) in \( (1, 2) \), it must be greater than or equal to the maximum possible value of \( -2x \), which is \( -2 \):
\( a \ge -2 \).
Therefore, the least value of \( a \) is \( -2 \).
In simple words: Differentiate to get \( 2x+a \). For the slope to stay positive on the interval, the constant \( a \) must be larger than \( -2x \). Since the largest value \( -2x \) approaches is -2, the minimum value for \( a \) is -2.

Exam Tip: Be sure to write the inequality steps clearly; when multiplying by a negative number (like -2), remember to reverse the inequality signs.

Short Answer Type Questions (4 Marks)

Question 56. Find the interval in which the function \( f(x) = 5x^{3/2} - 3x^{5/2} \), \( x > 0 \) is strictly decreasing.
Answer:
Differentiating the given function with respect to \( x \):
\( f'(x) = 5 \cdot \frac{3}{2}x^{1/2} - 3 \cdot \frac{5}{2}x^{3/2} = \frac{15}{2}\sqrt{x} - \frac{15}{2}x\sqrt{x} = \frac{15}{2}\sqrt{x}(1 - x) \).
For the function to be strictly decreasing, we must have \( f'(x) < 0 \):
\( \frac{15}{2}\sqrt{x}(1 - x) < 0 \).
Since \( x > 0 \), \( \sqrt{x} \) is always strictly positive. Thus, this inequality simplifies to:
\( 1 - x < 0 \implies x > 1 \).
Therefore, the interval where \( f(x) \) is strictly decreasing is \( (1, \infty) \).
In simple words: Find the derivative and factor it. Since \( x \) is positive, the sign of the slope depends solely on \( 1-x \). The slope is negative (decreasing) when \( x \) is greater than 1.

Exam Tip: Be sure to explicitly state that \( \sqrt{x} > 0 \) for \( x > 0 \) to justify why you can divide the inequality by \( \sqrt{x} \) without changing its direction.

Question 57. Show that the function \( f(x) = \tan^{-1}(\sin x + \cos x) \), is strictly increasing on the interval \( \left(0, \frac{\pi}{4}\right) \).
Answer:
Differentiating the function with respect to \( x \) using the chain rule:
\( f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx}(\sin x + \cos x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \xb \).
For \( f(x) \) to be strictly increasing, we need \( f'(x) > 0 \).
The denominator \( 1 + (\sin x + \cos x)^2 \) is a sum of squares and is always strictly positive.
So, the sign of \( f'(x) \) depends solely on the numerator \( \cos x - \sin x \).
In the given open interval \( x \in \left(0, \frac{\pi}{4}\right) \), we know that:
\( \cos x > \sin x \implies \cos x - \sin x > 0 \).
Thus, the derivative \( f'(x) \) is strictly positive on this interval.
Therefore, \( f(x) \) is strictly increasing on \( \left(0, \frac{\pi}{4}\right) \).
In simple words: Differentiate the function. Since the denominator is always positive, the sign of the slope depends on \( \cos x - \sin x \). This difference is positive between 0 and \( \pi/4 \) because cosine is larger than sine in this quadrant.

Exam Tip: Include a brief reference to the unit circle or trigonometric values to justify why \( \cos x > \sin x \) holds true on \( (0, \pi/4) \).

Question 58. Show that the function \( f(x) = \cos\left(2x + \frac{\pi}{4}\right) \) is strictly increasing on \( \left(\frac{3\pi}{8}, \frac{7\pi}{8}\right) \).
Answer:
Differentiating the function with respect to \( x \):
\( f'(x) = -\sin\left(2x + \frac{\pi}{4}\right) \cdot 2 = -2\sin\left(2x + \frac{\pi}{4}\right) \).
We are given the domain interval:
\( \frac{3\pi}{8} < x < \frac{7\pi}{8} \).
Multiply all parts of the inequality by 2:
\( \frac{3\pi}{4} < 2x < \frac{7\pi}{4} \).
Add \( \frac{\pi}{4} \):
\( \frac{3\pi}{4} + \frac{\pi}{4} < 2x + \frac{\pi}{4} < \frac{7\pi}{4} + \frac{\pi}{4} \)
\( \pi < 2x + \frac{\pi}{4} < 2\pi \).
Since the angle \( 2x + \frac{\pi}{4} \) lies strictly between \( \pi \) and \( 2\pi \) (the third and fourth quadrants), the sine of this angle is strictly negative:
\( \sin\left(2x + \frac{\pi}{4}\right) < 0 \).
Substituting this back into the derivative:
\( f'(x) = -2\sin\left(2x + \frac{\pi}{4}\right) > 0 \).
Since the derivative is strictly positive, the function is strictly increasing on \( \left(\frac{3\pi}{8}, \frac{7\pi}{8}\right) \).
In simple words: The derivative is \( -2\sin(2x+\pi/4) \). Scale the interval to show that the angle lies in the third and fourth quadrants where sine is negative. This negative sine makes the overall derivative positive, proving the function is increasing.

Exam Tip: Be precise when converting your inequalities; step-by-step scaling of intervals is a robust way to justify trigonometric sign behaviors.

Question 59. Show that the function \( f(x) = \frac{\sin x}{x} \) is strictly decreasing on \( \left(0, \frac{\pi}{2}\right) \).
Answer:
Differentiating the function using the quotient rule:
\( f'(x) = \frac{\cos x \cdot x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2} = \frac{\cos x (x - \tan x)}{x^2} \).
We analyze the signs of each term in the interval \( x \in \left(0, \frac{\pi}{2}\right) \):
- \( x^2 > 0 \) and \( \cos x > 0 \).
- For any acute angle in the first quadrant, the tangent is strictly larger than the angle itself: \( \tan x > x \implies x - \tan x < 0 \).
Thus, the numerator is strictly negative, which means:
\( f'(x) < 0 \) for all \( x \in \left(0, \frac{\pi}{2}\right) \).
Therefore, \( f(x) \) is strictly decreasing on \( \left(0, \frac{\pi}{2}\right) \).
In simple words: Differentiate the function using the quotient rule. Since \( \tan x > x \) for all acute angles, the numerator term \( x - \tan x \) is always negative, which makes the entire slope negative (decreasing).

Exam Tip: Stating the inequality \( \tan x > x \) for \( x \in (0, \pi/2) \) is a necessary justification step to prove that the derivative is negative.

Question 60. Using differentials, find the approximate value of \( (0.009)^{1/3} \).
Answer:
Let \( y = x^{1/3} \).
We choose the nearest perfect cube value \( x = 0.008 \) (so \( y = 0.008^{1/3} = 0.2 \)) and the difference \( \Delta x = 0.001 \).
Using differentials to calculate the small change \( dy \):
\( dy = \frac{dy}{dx} \Delta x = \frac{1}{3} x^{-2/3} \Delta x \).
Substitute the chosen values:
\( dy = \frac{1}{3} (0.008)^{-2/3} (0.001) = \frac{1}{3} \left[(0.2)^3\right]^{-2/3} (0.001) = \frac{1}{3} (0.04)^{-1} (0.001) \)
\( = \frac{1}{3 \times 0.04} (0.001) = \frac{0.001}{0.12} = \frac{1}{120} \approx 0.0083 \).
Now approximate the function:
\( (0.009)^{1/3} \approx y + dy = 0.2 + 0.0083 = 0.2083 \).
Therefore, the approximate value is \( 0.2083 \).
In simple words: Choose \( x = 0.008 \) as a perfect cube reference point. Use calculus to find the small difference \( dy \) for the step of \( 0.001 \), and add it to \( 0.2 \) to find the approximation.

Exam Tip: Always state your choice of \( x \) and \( \Delta x \) clearly at the beginning of your calculation; this is expected for full scoring.

Question 61. Using differentials, find the approximate value of \( (255)^{1/4} \).
Answer:
Let \( y = x^{1/4} \).
We choose the nearest perfect fourth-power value \( x = 256 \) (so \( y = 256^{1/4} = 4 \)) and the difference \( \Delta x = -1 \xb \).
Using differentials to find \( dy \):
\( dy = \frac{dy}{dx} \Delta x = \frac{1}{4} x^{-3/4} \Delta x \).
Substitute the chosen values:
\( dy = \frac{1}{4} (256)^{-3/4} (-1) = \frac{-1}{4 \times \left(256^{1/4}\right)^3} = \frac{-1}{4 \times 4^3} = -\frac{1}{256} \approx -0.0039 \).
Now calculate the approximation:
\( (255)^{1/4} \approx y + dy = 4 - 0.0039 = 3.9961 \).
Therefore, the approximate value is \( 3.9961 \).
In simple words: Use \( 256 \) as your reference point since \( 256^{1/4} = 4 \). Calculate the negative difference \( dy \) for the step of \( -1 \) using derivatives, and subtract it from 4 to find the approximation.

Exam Tip: When \( \Delta x \) is negative, ensure your differential \( dy \) is also negative so that the approximated value is correctly less than the reference value.

Question 62. Using differentials, find the approximate value of \( (0.0037)^{1/2} \).
Answer:
Let \( y = \sqrt{x} \).
We choose the nearest perfect square value \( x = 0.0036 \) (so \( y = \sqrt{0.0036} = 0.06 \)) and the difference \( \Delta x = 0.0001 \).
Using differentials to find the change \( dy \):
\( dy = \frac{dy}{dx} \Delta x = \frac{1}{2\sqrt{x}} \Delta x \).
Substitute the chosen values:
\( dy = \frac{1}{2\sqrt{0.0036}} (0.0001) = \frac{0.0001}{2(0.06)} = \frac{0.0001}{0.12} = \frac{1}{1200} \approx 0.00083 \).
Now approximate:
\( (0.0037)^{1/2} \approx y + dy = 0.06 + 0.00083 = 0.06083 \).
Therefore, the approximate value is \( 0.06083 \).
In simple words: Use \( 0.0036 \) as a perfect square reference. Find the tiny change \( dy \) using calculus for the step of \( 0.0001 \), and add it to \( 0.06 \) to get the approximation.

Exam Tip: Be careful with decimals during the division step; writing them out in fraction form first can prevent decimal point placement errors.

Question 63. Using differentials, find the approximate value of \( \sqrt{0.037} \).
Answer:
Let \( y = \sqrt{x} \).
We choose the nearest perfect square value \( x = 0.04 \) (so \( y = \sqrt{0.04} = 0.2 \)) and the difference \( \Delta x = -0.003 \).
Using differentials to find \( dy \):
\( dy = \frac{dy}{dx} \Delta x = \frac{1}{2\sqrt{x}} \Delta x \).
Substitute the chosen values:
\( dy = \frac{1}{2\sqrt{0.04}} (-0.003) = \frac{-0.003}{2(0.2)} = -\frac{0.003}{0.4} = -0.0075 \).
Now approximate:
\( \sqrt{0.037} \approx y + dy = 0.2 - 0.0075 = 0.1925 \).
Therefore, the approximate value is \( 0.1925 \).
In simple words: Use \( 0.04 \) as your perfect square reference. Find the negative differential \( dy \) for the step of \( -0.003 \) using derivatives, and subtract it from 0.2 to get the final answer.

Exam Tip: Make sure you perform the subtraction \( 0.2 - 0.0075 \) carefully to avoid basic arithmetic slips on the final step.

Question 64. Using differentials, find the approximate value of \( \sqrt{25.3} \).
Answer:
Let \( y = \sqrt{x} \).
We choose the perfect square \( x = 25 \) (so \( y = \sqrt{25} = 5 \)) and the difference \( \Delta x = 0.3 \).
Using differentials to calculate the small change \( dy \):
\( dy = \frac{dy}{dx} \Delta x = \frac{1}{2\sqrt{x}} \Delta x \).
Substitute the values:
\( dy = \frac{1}{2\sqrt{25}} (0.3) = \frac{0.3}{10} = 0.03 \).
Now calculate the approximation:
\( \sqrt{25.3} \approx y + dy = 5 + 0.03 = 5.03 \).
Therefore, the approximate value is \( 5.03 \).
In simple words: Choose \( 25 \) as a reference point. Calculate the small change \( dy \) using derivatives for the step of \( 0.3 \), and add it to 5 to get the approximation.

Exam Tip: This is a classic board exam question. Its calculations are extremely straightforward, making it an easy source of full marks.

Question 65. Find the approximate value of f (5.001) where f(x) = \( x^3 \) – 7\( x^2 \) + 15.
Answer:
We use the linear approximation formula:
\( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \).
We choose the reference point \( x = 5 \) and the small step \( \Delta x = 0.001 \).
First, calculate the value of the function at \( x = 5 \):
\( f(5) = 5^3 - 7(5)^2 + 15 = 125 - 175 + 15 = -35 \).
Now, differentiate the function:
\( f'(x) = 3x^2 - 14x \).
Evaluate the derivative at \( x = 5 \):
\( f'(5) = 3(5)^2 - 14(5) = 75 - 70 = 5 \).
Now substitute these values into the approximation formula:
\( f(5.001) \approx f(5) + f'(5) \Delta x = -35 + 5(0.001) = -35 + 0.005 = -34.995 \).
Therefore, the approximate value is \( -34.995 \).
In simple words: Calculate the function value at \( x = 5 \) (which is -35). Find the derivative at \( x = 5 \) (which is 5), multiply it by the small step \( 0.001 \), and add it to find the approximation.

Exam Tip: Write down the approximation formula \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \) at the beginning to establish a clear mathematical basis for your steps.

Question 66. Find the approximate value of f (3.02) where f (x) = 3\( x^2 \) + 5x + 3.
Answer:
We use the linear approximation formula:
\( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \).
We choose the reference point \( x = 3 \) and the small step \( \Delta x = 0.02 \).
First, calculate the value of the function at \( x = 3 \):
\( f(3) = 3(3)^2 + 5(3) + 3 = 27 + 15 + 3 = 45 \).
Now, differentiate the function:
\( f'(x) = 6x + 5 \).
Evaluate the derivative at \( x = 3 \):
\( f'(3) = 6(3) + 5 = 23 \).
Substitute these values into the approximation formula:
\( f(3.02) \approx f(3) + f'(3) \Delta x = 45 + 23(0.02) = 45 + 0.46 = 45.46 \).
Therefore, the approximate value is \( 45.46 \).
In simple words: Calculate the function value at \( x = 3 \) (which is 45). Differentiate to find the slope at \( x = 3 \) (which is 23), multiply by the step of \( 0.02 \), and add them together.

Exam Tip: Double check your addition at the end; simple arithmetic slips on the final step can cost you easy marks.

Long Answer Type Questions (6 Marks)

Question 67. Show that of all rectangles inscribed in a given fixed circle, the square has the maximum area.
Answer:
Let the given fixed circle have radius \( R \).
Let a rectangle inscribed in this circle have length \( x \) and width \( y \).
By the Pythagorean theorem, the diagonal of the rectangle is equal to the diameter of the circle:
\( x^2 + y^2 = (2R)^2 = 4R^2 \)

\( \implies y = \sqrt{4R^2 - x^2} \).
The area \( A \) of the rectangle is:
\( A = x \cdot y = x \sqrt{4R^2 - x^2} \).
To maximize the area \( A \), we can maximize its square \( Z = A^2 \):
\( Z = x^2 (4R^2 - x^2) = 4R^2 x^2 - x^4 \).
Differentiating \( Z \) with respect to \( x \):
\( \frac{dZ}{dx} = 8R^2 x - 4x^3 \).
Setting this first derivative to zero to find the critical points:
\( 4x(2R^2 - x^2) = 0 \).
Since \( x > 0 \), we have:
\( 2R^2 - x^2 = 0 \implies x^2 = 2R^2 \implies x = \sqrt{2}R \).
Find the corresponding width \( y \) at this value of \( x \):
\( y = \sqrt{4R^2 - x^2} = \sqrt{4R^2 - 2R^2} = \sqrt{2R^2} = \sqrt{2}R \).
Since the length is equal to the width (\( x = y = \sqrt{2}R \)), the rectangle is a square.

Now, use the second derivative test to confirm it is a maximum:
\( \frac{d^2 Z}{dx^2} = \frac{d}{dx}(8R^2 x - 4x^3) = 8R^2 - 12x^2 \).
At \( x = \sqrt{2}R \):
\( \frac{d^2 Z}{dx^2} = 8R^2 - 12(2R^2) = 8R^2 - 24R^2 = -16R^2 < 0 \xb \).
Since the second derivative is negative, the area is a maximum.
Hence proved.
In simple words: Relate the sides of the rectangle to the circle's diameter using Pythagoras. Write down the area formula, differentiate to find where the slope is 0, show the length and width are equal (making it a square), and use the second derivative test to confirm it is a maximum.

Exam Tip: Maximizing the squared area \( A^2 \) instead of \( A \) is a highly useful trick that eliminates the square root, making differentiation much simpler.

Question 68. Find two positive numbers x and y such that their sum is 35 and the product \( x^2y^5 \) is maximum.
Answer:
We are given that the sum of the two positive numbers is 35:
\( x + y = 35 \implies x = 35 - y \).
Let the product be \( P = x^2 y^5 = (35 - y)^2 y^5 \).
Taking the natural logarithm on both sides to simplify differentiation:
\( \ln P = 2 \ln(35 - y) + 5 \ln y \xb \).
Differentiating both sides with respect to \( y \):
\( \frac{1}{P} \frac{dP}{dy} = \frac{-2}{35 - y} + \frac{5}{y} = \frac{-2y + 5(35 - y)}{y(35 - y)} = \frac{175 - 7y}{y(35 - y)} \).
Setting the derivative to zero to find the critical points:
\( \frac{dP}{dy} = 0 \implies 175 - 7y = 0 \implies 7y = 175 \implies y = 25 \).
Find the value of \( x \) when \( y = 25 \):
\( x = 35 - 25 = 10 \).

Now, use the second derivative test to confirm it is a maximum:
\( \frac{d}{dy}\left(\frac{1}{P}\frac{dP}{dy}\right) = \frac{-7[y(35-y)] - (175-7y)(35-2y)}{[y(35-y)]^2} \).
At \( y = 25 \) (where \( 175 - 7y = 0 \)):
\( = \frac{-7(25)(10)}{[25(10)]^2} = -\frac{1750}{62500} < 0 \).
Since the second derivative is negative, the product is indeed a maximum.
Therefore, the required positive numbers are \( x = 10 \) and \( y = 25 \).
In simple words: Write \( x \) in terms of \( y \), substitute into the product, use logs to differentiate easily, find where the derivative is 0 to get \( y=25 \), and confirm it is a maximum using second derivatives.

Exam Tip: Logarithmic differentiation is highly recommended for products with large power terms like \( x^2 y^5 \); it saves significant algebraic factorization steps.

Question 69. Show that of all the rectangles of given area, the square has the smallest perimeter.
Answer:
Let the given constant area of the rectangle be \( A \).
Let the dimensions of the rectangle be \( x \) and \( y \), so:
\( x \cdot y = A \implies y = \frac{A}{x} \).
The perimeter \( P \) of the rectangle is given by:
\( P = 2(x + y) = 2\left(x + \frac{A}{x}\right) \).
Differentiating \( P \) with respect to \( x \):
\( \frac{dP}{dx} = 2\left(1 - \frac{A}{x^2}\right) \).
Setting the derivative to zero to find the critical points:
\( 2\left(1 - \frac{A}{x^2}\right) = 0 \implies 1 = \frac{A}{x^2} \implies x^2 = A \implies x = \sqrt{A} \) (since \( x > 0 \)).
Find the corresponding width \( y \) at this value of \( x \):
\( y = \frac{A}{\sqrt{A}} = \sqrt{A} \).
Since the length is equal to the width (\( x = y = \sqrt{A} \)), the rectangle is a square.

Now, use the second derivative test to confirm it is a minimum:
\( \frac{d^2 P}{dx^2} = 2\left(0 - \left(-\frac{2A}{x^3}\right)\right) = \frac{4A}{x^3} \).
Since \( A > 0 \) and \( x > 0 \), the second derivative is strictly positive:
\( \frac{d^2 P}{dx^2} > 0 \).
Since the second derivative is positive, the perimeter is a minimum.
Hence proved.
In simple words: Write the perimeter formula in terms of one variable \( x \) by using the area constant. Differentiate, set to 0 to show length equals width (square), and use second derivatives to confirm it is the smallest perimeter.

Exam Tip: Be sure to write "constant area" at the start, as identifying \( A \) as a constant parameter is essential for correct differentiation.

Question 70. Show that the right circular cone of least curved surface area and given volume has an altitude equal to \( \sqrt{2} \) times the radium of the base.
Answer:
Let the given volume of the cone be a constant \( V \).
Let the cone have radius \( r \), height (altitude) \( h \), and slant height \( l = \sqrt{r^2 + h^2} \).
The volume \( V \) of a cone is given by:
\( V = \frac{1}{3}\pi r^2 h \implies h = \frac{3V}{\pi r^2} \).
The curved surface area \( S \) of the cone is:
\( S = \pi r l = \pi r \sqrt{r^2 + h^2} \).
Squaring \( S \) to simplify differentiation:
\( Z = S^2 = \pi^2 r^2 (r^2 + h^2) \).
Substituting the value of \( h \) in terms of \( r \):
\( Z = \pi^2 r^2 \left(r^2 + \frac{9V^2}{\pi^2 r^4}\right) = \pi^2 r^4 + \frac{9V^2}{r^2} \).
Differentiating \( Z \) with respect to the radius \( r \):
\( \frac{dZ}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3} \).
Setting this first derivative to zero to find the critical points:
\( 4\pi^2 r^3 = \frac{18V^2}{r^3} \implies r^6 = \frac{18V^2}{4\pi^2} = \frac{9V^2}{2\pi^2} \).
Substitute \( V = \frac{1}{3}\pi r^2 h \) back into the equation:
\( r^6 = \frac{9\left(\frac{1}{3}\pi r^2 h\right)^2}{2\pi^2} = \frac{\pi^2 r^4 h^2}{2\pi^2} = \frac{r^4 h^2}{2} \).
Dividing both sides by the positive term \( r^4 \):
\( r^2 = \frac{h^2}{2} \implies h^2 = 2r^2 \implies h = \sqrt{2}r \).
Thus, the altitude \( h \) is equal to \( \sqrt{2} \) times the radius of the base \( r \).

Now, use the second derivative test to confirm it is a minimum:
\( \frac{d^2 Z}{dr^2} = 12\pi^2 r^2 + \frac{54V^2}{r^4} \).
Since \( r > 0 \) and \( V^2 > 0 \), the second derivative is strictly positive:
\( \frac{d^2 Z}{dr^2} > 0 \).
Since the second derivative is positive, the curved surface area is a minimum.
Hence proved.
In simple words: Express height in terms of radius using the volume constant. Write the squared curved surface area, differentiate with respect to \( r \), set to 0, substitute the volume expression back to simplify, and prove \( h = \sqrt{2}r \).

Exam Tip: Substituting the volume \( V \) back into the solved \( r^6 \) expression at the end is a highly elegant approach that simplifies the proof of the ratio \( h = \sqrt{2}r \).

Question 71. Show that the semi vertical angle of right circular cone of given surface area and maximum volume is \( \sin^{-1}\left(\frac{1}{3}\right) \).
Answer: Let \( r \) be the radius, \( h \) be the height, and \( l \) be the slant height of the cone.
The total surface area \( S \) of the cone is given as:
\( S = \pi r^2 + \pi r l \)
From this, we can express the slant height \( l \) in terms of \( S \) and \( r \):
\( l = \frac{S - \pi r^2}{\pi r} \)
Since the height \( h \), radius \( r \), and slant height \( l \) form a right-angled triangle, we have:
\( h^2 = l^2 - r^2 \)
Substituting the value of \( l \):
\( h^2 = \left(\frac{S - \pi r^2}{\pi r}\right)^2 - r^2 \)

\( h^2 = \frac{(S - \pi r^2)^2 - \pi^2 r^4}{\pi^2 r^2} \)

\( h^2 = \frac{S^2 - 2S\pi r^2}{\pi^2 r^2} \)

\( h^2 = \frac{S(S - 2\pi r^2)}{\pi^2 r^2} \)
The volume \( V \) of the cone is:
\( V = \frac{1}{3} \pi r^2 h \)
Let us square the volume to simplify the differentiation:
\( Z = V^2 = \frac{1}{9} \pi^2 r^4 h^2 \)
Substitute the expression for \( h^2 \):
\( Z = \frac{1}{9} \pi^2 r^4 \cdot \frac{S(S - 2\pi r^2)}{\pi^2 r^2} \)

\( Z = \frac{S}{9} r^2 (S - 2\pi r^2) \)

\( Z = \frac{S^2 r^2 - 2S\pi r^4}{9} \)
Now, we differentiate \( Z \) with respect to \( r \):
\( \frac{dZ}{dr} = \frac{1}{9} [2 S^2 r - 8 S \pi r^3] \)
For maximum volume, we set \( \frac{dZ}{dr} = 0 \):
\( 2 S^2 r - 8 S \pi r^3 = 0 \)
Since \( S \neq 0 \) and \( r \neq 0 \):
\( 2S = 8\pi r^2 \)

\( \implies S = 4\pi r^2 \)
Substitute \( S = \pi r^2 + \pi r l \) into this equation:
\( \pi r^2 + \pi r l = 4\pi r^2 \)

\( \implies \pi r l = 3\pi r^2 \)

\( \implies l = 3r \)

\( \implies \frac{r}{l} = \frac{1}{3} \)
Let \( \theta \) be the semi-vertical angle of the cone. Then, \( \sin\theta = \frac{r}{l} = \frac{1}{3} \).
Thus, the semi-vertical angle is:
\( \theta = \sin^{-1}\left(\frac{1}{3}\right) \)
To verify that this point yields the maximum volume, we take the second derivative:
\( \frac{d^2Z}{dr^2} = \frac{1}{9} [2 S^2 - 24 S \pi r^2] \)
Substituting \( S = 4\pi r^2 \):
\( \frac{d^2Z}{dr^2} = \frac{1}{9} [2(4\pi r^2)^2 - 24(4\pi r^2)\pi r^2] \)

\( \frac{d^2Z}{dr^2} = \frac{1}{9} [32\pi^2 r^4 - 96\pi^2 r^4] = -\frac{64}{9}\pi^2 r^4 < 0 \)
Since the second derivative is negative, the volume is indeed maximized at this angle.
In simple words: When the surface area of a cone is fixed, the largest possible volume is achieved when the ratio of the radius to the slant height is exactly 1 to 3. This corresponds to a semi-vertical angle of \( \sin^{-1}(1/3) \).

Exam Tip: Squaring the volume function before differentiating is a highly effective technique to avoid dealing with complicated square root derivatives.

Question 72. A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is \( \left(a^{2/3} + b^{2/3}\right)^{3/2} \).
Answer: Let \( ABC \) be a right-angled triangle, right-angled at \( B \). Let \( P \) be a point on the hypotenuse \( AC \) such that the perpendicular distance of \( P \) from \( AB \) is \( b \) and from \( BC \) is \( a \).
Let the hypotenuse \( AC \) make an angle \( \theta \) with the side \( BC \).
From the geometry of the triangle, we have:
\( AP = b \sec\theta \) and \( PC = a \csc\theta \)
The total length of the hypotenuse \( L \) is:
\( L = AP + PC = b \sec\theta + a \csc\theta \)
To find the minimum length, we differentiate \( L \) with respect to \( \theta \):
\( \frac{dL}{d\theta} = b \sec\theta \tan\theta - a \csc\theta \cot\theta \)
Setting the first derivative to zero for critical points:
\( b \sec\theta \tan\theta - a \csc\theta \cot\theta = 0 \)

\( \implies \frac{b \sin\theta}{\cos^2\theta} = \frac{a \cos\theta}{\sin^2\theta} \)

\( \implies \tan^3\theta = \frac{a}{b} \)

\( \implies \tan\theta = \left(\frac{a}{b}\right)^{1/3} \)
Using this relation, we can find the values of \( \sec\theta \) and \( \csc\theta \):
\( \sec\theta = \frac{\sqrt{a^{2/3} + b^{2/3}}}{b^{1/3}} \)
\( \csc\theta = \frac{\sqrt{a^{2/3} + b^{2/3}}}{a^{1/3}} \)
Substituting these back into the length equation:
\( L = b \left(\frac{\sqrt{a^{2/3} + b^{2/3}}}{b^{1/3}}\right) + a \left(\frac{\sqrt{a^{2/3} + b^{2/3}}}{a^{1/3}}\right) \)

\( L = b^{2/3}\sqrt{a^{2/3} + b^{2/3}} + a^{2/3}\sqrt{a^{2/3} + b^{2/3}} \)

\( L = \left(a^{2/3} + b^{2/3}\right)\sqrt{a^{2/3} + b^{2/3}} \)

\( L = \left(a^{2/3} + b^{2/3}\right)^{3/2} \)
Since \( \frac{d^2L}{d\theta^2} > 0 \) for acute angle \( \theta \), this represents the absolute minimum length of the hypotenuse.
In simple words: By defining the position of the point using trigonometry, we can construct a function for the length of the hypotenuse. Differentiating this function shows that the shortest hypotenuse that can pass through that point is \( (a^{2/3} + b^{2/3})^{3/2} \).

Exam Tip: Expressing the lengths of the segments using a single angle \( \theta \) simplifies the optimization problem considerably compared to using Cartesian coordinates.

Question 73. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \( \frac{8}{27} \) of the volume of the sphere.
Answer: Let \( R \) be the radius of the sphere with center \( O \). Let a cone of radius \( r \) and height \( h \) be inscribed in this sphere.
Let \( x \) be the distance from the center of the sphere \( O \) to the base of the cone.
The height of the cone is given by:
\( h = R + x \)
In the right-angled triangle formed by the radius of the sphere, the radius of the cone base, and the distance \( x \), we have:
\( r^2 = R^2 - x^2 \)
The volume \( V \) of the inscribed cone is:
\( V = \frac{1}{3} \pi r^2 h \)
Substitute the value of \( r^2 \) and \( h \):
\( V = \frac{1}{3} \pi (R^2 - x^2)(R + x) \)
Now, we differentiate \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{\pi}{3} \left[ (R^2 - x^2)(1) + (R + x)(-2x) \right] \)

\( \frac{dV}{dx} = \frac{\pi}{3} (R + x) \left[ (R - x) - 2x \right] \)

\( \frac{dV}{dx} = \frac{\pi}{3} (R + x)(R - 3x) \)
Setting \( \frac{dV}{dx} = 0 \), we obtain critical points:
\( x = -R \) (not possible as it makes height zero) or \( x = \frac{R}{3} \).
Let's compute the second derivative to verify the maximum:
\( \frac{d^2V}{dx^2} = \frac{\pi}{3} [ (R - 3x) + (R + x)(-3) ] = \frac{\pi}{3} [ -2R - 6x ] \)
At \( x = \frac{R}{3} \):
\( \frac{d^2V}{dx^2} = \frac{\pi}{3} [ -2R - 2R ] = -\frac{4\pi R}{3} < 0 \)
Thus, the volume is maximized at \( x = \frac{R}{3} \).
The maximum volume of the cone is:
\( V_{\text{max}} = \frac{1}{3} \pi \left( R^2 - \frac{R^2}{9} \right) \left( R + \frac{R}{3} \right) = \frac{1}{3} \pi \left( \frac{8R^2}{9} \right) \left( \frac{4R}{3} \right) = \frac{32}{81} \pi R^3 \)
The volume of the sphere is:
\( V_{\text{sphere}} = \frac{4}{3} \pi R^3 \)
Comparing the two volumes:
\( V_{\text{max}} = \frac{32}{81} \pi R^3 = \frac{8}{27} \left( \frac{4}{3} \pi R^3 \right) = \frac{8}{27} V_{\text{sphere}} \)
Hence proved.
In simple words: The largest cone you can fit inside a sphere will always take up exactly \( \frac{8}{27} \) (or about 29.6%) of the sphere's total volume. This occurs when the base of the cone is positioned at a distance of one-third of the radius below the center.

Exam Tip: Be sure to write the final step showing the exact factor of \( \frac{8}{27} \) multiplied by the standard formula of the volume of a sphere to get full presentation marks.

Question 74. Find the interval in which the function f given by f(x) = sin x + cos x, 0 ≤ x ≤ 2π is strictly increasing or strictly decreasing.
Answer: The given function is:
\( f(x) = \sin x + \cos x \)
Differentiating with respect to \( x \):
\( f'(x) = \cos x - \sin x \)
To find the critical points, we set \( f'(x) = 0 \):
\( \cos x - \sin x = 0 \)

\( \implies \tan x = 1 \)
Since \( 0 \le x \le 2\pi \), the solution points are:
\( x = \frac{\pi}{4}, \frac{5\pi}{4} \)
These points partition the domain \( [0, 2\pi] \) into three subintervals:
1. \( \left[0, \frac{\pi}{4}\right) \):
For any point in this interval, say \( x = 0 \), we have \( f'(0) = \cos 0 - \sin 0 = 1 > 0 \). Thus, \( f(x) \) is strictly increasing in \( \left[0, \frac{\pi}{4}\right) \).
2. \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \):
For any point in this interval, say \( x = \pi \), we have \( f'(\pi) = \cos\pi - \sin\pi = -1 < 0 \). Thus, \( f(x) \) is strictly decreasing in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).
3. \( \left(\frac{5\pi}{4}, 2\pi\right] \):
For any point in this interval, say \( x = \frac{3\pi}{2} \), we have \( f'\left(\frac{3\pi}{2}\right) = \cos\frac{3\pi}{2} - \sin\frac{3\pi}{2} = 1 > 0 \). Thus, \( f(x) \) is strictly increasing in \( \left(\frac{5\pi}{4}, 2\pi\right] \).
Consequently:
- The function is strictly increasing in \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \).
- The function is strictly decreasing in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).
In simple words: The derivative of the function shows whether it goes up or down. For this function, the values rise in the beginning and at the end of the \( 2\pi \) cycle, but fall in the middle section between \( \pi/4 \) and \( 5\pi/4 \).

Exam Tip: Clearly show the test value calculation for each interval so the examiner can see how you determined the sign of the derivative.

Question 75. Find the intervals in which the function f(x) = (x + 1)³ (x - 3)³ is strictly increasing or strictly decreasing.
Answer: The function is defined as:
\( f(x) = (x + 1)^3 (x - 3)^3 \)
Differentiating \( f(x) \) using the product rule:
\( f'(x) = 3(x + 1)^2 (x - 3)^3 + 3(x + 1)^3 (x - 3)^2 \)
Factoring out the common terms:
\( f'(x) = 3(x + 1)^2 (x - 3)^2 [ (x - 3) + (x + 1) ] \)

\( f'(x) = 3(x + 1)^2 (x - 3)^2 (2x - 2) \)

\( f'(x) = 6(x + 1)^2 (x - 3)^2 (x - 1) \)
To find critical points, we set \( f'(x) = 0 \), yielding:
\( x = -1, 1, 3 \)
Note that \( (x + 1)^2 \) and \( (x - 3)^2 \) are non-negative for all real \( x \). Therefore, the sign of \( f'(x) \) is entirely determined by the factor \( (x - 1) \):
- If \( x < 1 \) (and \( x \neq -1 \)), then \( x - 1 < 0 \), which makes \( f'(x) < 0 \).
- If \( x > 1 \) (and \( x \neq 3 \)), then \( x - 1 > 0 \), which makes \( f'(x) > 0 \).
Therefore, the intervals are:
- Strictly decreasing in \( (-\infty, 1) \).
- Strictly increasing in \( (1, \infty) \).
In simple words: Since the squared terms in the derivative are always positive, only the term \( (x - 1) \) changes the overall sign. The function slopes downward for any value less than 1, and slopes upward for any value greater than 1.

Exam Tip: Since squares of real numbers are always positive, you can simplify the sign-chart analysis by focusing solely on the sign of the term with the odd power.

Question 76. Find the local maximum and local minimum of f(x) = sin 2x - x, -\frac{\pi}{2} < x < \frac{\pi}{2}.
Answer: The function is:
\( f(x) = \sin 2x - x \)
Differentiating with respect to \( x \):
\( f'(x) = 2\cos 2x - 1 \)
For local maxima or minima, we set \( f'(x) = 0 \):
\( 2\cos 2x - 1 = 0 \)

\( \implies \cos 2x = \frac{1}{2} \)
Since \( -\frac{\pi}{2} < x < \frac{\pi}{2} \), we have \( -\pi < 2x < \pi \). In this interval, the solutions for \( 2x \) are:
\( 2x = -\frac{\pi}{3}, \frac{\pi}{3} \)

\( \implies x = -\frac{\pi}{6}, \frac{\pi}{6} \)
Now, let's find the second derivative of \( f(x) \):
\( f''(x) = -4\sin 2x \)
Evaluating at the critical points:
- At \( x = \frac{\pi}{6} \):
\( f''\left(\frac{\pi}{6}\right) = -4\sin\left(\frac{\pi}{3}\right) = -4\left(\frac{\sqrt{3}}{2}\right) = -2\sqrt{3} < 0 \)
Since the second derivative is negative, \( x = \frac{\pi}{6} \) is a point of local maximum.
The local maximum value is:
\( f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \)

- At \( x = -\frac{\pi}{6} \):
\( f''\left(-\frac{\pi}{6}\right) = -4\sin\left(-\frac{\pi}{3}\right) = 2\sqrt{3} > 0 \)
Since the second derivative is positive, \( x = -\frac{\pi}{6} \) is a point of local minimum.
The local minimum value is:
\( f\left(-\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{3}\right) - \left(-\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{\pi}{6} \)
In simple words: Setting the derivative to zero gives two key points. Evaluating the second derivative tells us that \( x = \pi/6 \) is the peak (local maximum) and \( x = -\pi/6 \) is the valley (local minimum) of this curve in the given region.

Exam Tip: Be careful with the domain transformation when going from \( x \) to \( 2x \). This ensures you do not miss or add incorrect critical points.

Question 77. Find the intervals in which the function f(x) = 2x³ - 15x² + 36x + 1 is strictly increasing or decreasing. Also find the points on which the tangents are parallel to x-axis.
Answer: The given function is:
\( f(x) = 2x^3 - 15x^2 + 36x + 1 \)
Differentiating with respect to \( x \):
\( f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3) \)
To find critical points, set \( f'(x) = 0 \):
\( 6(x - 2)(x - 3) = 0 \)

\( \implies x = 2, 3 \)
These points divide the real line into three intervals: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \).
1. In \( (-\infty, 2) \): \( f'(x) > 0 \) (strictly increasing).
2. In \( (2, 3) \): \( f'(x) < 0 \) (strictly decreasing).
3. In \( (3, \infty) \): \( f'(x) > 0 \) (strictly increasing).

Thus:
- Strictly increasing in \( (-\infty, 2) \cup (3, \infty) \).
- Strictly decreasing in \( (2, 3) \).

The tangents are parallel to the x-axis where the slope \( f'(x) = 0 \), which occurs at \( x = 2 \) and \( x = 3 \).
- For \( x = 2 \):
\( y = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29 \)
The point is \( (2, 29) \).

- For \( x = 3 \):
\( y = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28 \)
The point is \( (3, 28) \).
In simple words: The curve climbs up until \( x = 2 \), goes down between \( x = 2 \) and \( x = 3 \), and then climbs up again. The tangent lines are perfectly flat (parallel to the x-axis) at the turning points, which are \( (2, 29) \) and \( (3, 28) \).

Exam Tip: Remember to substitute the \( x \)-values back into the original function \( f(x) \) (not the derivative) to find the correct coordinates for the points of tangency.

Question 78. A solid is formed by a cylinder of radius r and height h together with two hemisphere of radius r attached at each end. It the volume of the solid is constant but radius r is increasing at the rate of \( \frac{1}{2\pi} \) metre/min. How fast must h (height) be changing when r and h are 10 metres.
Answer: Let \( V \) be the total volume of the solid. The solid consists of a cylinder of radius \( r \) and height \( h \), and two hemispheres of radius \( r \).
The volume \( V \) is given by:
\( V = \pi r^2 h + 2\left(\frac{2}{3}\pi r^3\right) = \pi r^2 h + \frac{4}{3}\pi r^3 \)
Since the volume \( V \) is constant, its rate of change with respect to time \( t \) is zero:
\( \frac{dV}{dt} = 0 \)
Differentiating the volume expression with respect to \( t \):
\( \frac{dV}{dt} = \pi \left[ r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} \right] + 4\pi r^2 \frac{dr}{dt} = 0 \)
Dividing the entire equation by \( \pi r \) (since \( r \neq 0 \)):
\( r \frac{dh}{dt} + 2h \frac{dr}{dt} + 4r \frac{dr}{dt} = 0 \)

\( \implies r \frac{dh}{dt} = -(2h + 4r)\frac{dr}{dt} \)

\( \implies \frac{dh}{dt} = -\frac{2h + 4r}{r} \frac{dr}{dt} \)
We are given \( r = 10 \), \( h = 10 \), and \( \frac{dr}{dt} = \frac{1}{2\pi} \). Substituting these values into the expression:
\( \frac{dh}{dt} = -\frac{2(10) + 4(10)}{10} \left( \frac{1}{2\pi} \right) \)

\( \frac{dh}{dt} = -\frac{60}{10} \left( \frac{1}{2\pi} \right) = -6 \left( \frac{1}{2\pi} \right) = -\frac{3}{\pi} \text{ metres/min} \)
Therefore, the height \( h \) is decreasing at the rate of \( \frac{3}{\pi} \) metres/min.
In simple words: Since the total volume of the shape cannot change, as the radius of the cylinder and hemispheres grows, the height of the cylinder must shrink. At the given moment, the height is decreasing at a rate of \( 3/\pi \) metres per minute.

Exam Tip: The negative sign in your final answer indicates a decrease. Make sure to clearly state that the height is decreasing when writing your final sentence.

Question 79. Find the equation of the normal to the curve x = a (cos θ + θ sin θ) ; y = a (sin θ - θ cos θ) at the point θ and show that its distance from the origin is a.
Answer: The parametric equations of the curve are:
\( x = a(\cos\theta + \theta\sin\theta) \)
\( y = a(\sin\theta - \theta\cos\theta) \)
Differentiating both with respect to \( \theta \):
\( \frac{dx}{d\theta} = a(-\sin\theta + \sin\theta + \theta\cos\theta) = a\theta\cos\theta \)
\( \frac{dy}{d\theta} = a(\cos\theta - \cos\theta + \theta\sin\theta) = a\theta\sin\theta \)
The slope of the tangent to the curve is:
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta \)
The slope of the normal is the negative reciprocal of the tangent's slope:
\( m_n = -\frac{1}{\tan\theta} = -\frac{\cos\theta}{\sin\theta} \)
The equation of the normal at the point \( \theta \) is:
\( y - y_0 = m_n(x - x_0) \)
\( y - a(\sin\theta - \theta\cos\theta) = -\frac{\cos\theta}{\sin\theta} [ x - a(\cos\theta + \theta\sin\theta) ] \)
Multiplying both sides by \( \sin\theta \):
\( y\sin\theta - a\sin^2\theta + a\theta\sin\theta\cos\theta = -x\cos\theta + a\cos^2\theta + a\theta\sin\theta\cos\theta \)
Simplifying by cancelling the common term \( a\theta\sin\theta\cos\theta \) on both sides:
\( x\cos\theta + y\sin\theta = a(\cos^2\theta + \sin^2\theta) \)

\( \implies x\cos\theta + y\sin\theta - a = 0 \)
The perpendicular distance \( d \) of the normal line \( A x + B y + C = 0 \) from the origin \( (0, 0) \) is given by:
\( d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|-a|}{\sqrt{\cos^2\theta + \sin^2\theta}} = \frac{a}{1} = a \)
Hence, the distance of the normal from the origin is always \( a \).
In simple words: By finding the derivatives of the parametric coordinates, we calculate the slope of the normal line. Working out the line equation simplifies it to a neat form, and using the distance formula shows that it is always at a distance of exactly \( a \) from the origin, regardless of the angle.

Exam Tip: Be meticulous with the product rule when differentiating the term \( \theta\sin\theta \) and \( \theta\cos\theta \); sign errors here are very common.

Question 80. For the curve y = 4x³ - 2x⁵, find all the points at which the tangent passes through the origin.
Answer: Let \( (x_1, y_1) \) be a point of contact on the given curve:
\( y_1 = 4x_1^3 - 2x_1^5 \)
The derivative of the curve with respect to \( x \) is:
\( \frac{dy}{dx} = 12x^2 - 10x^4 \)
Thus, the slope of the tangent at \( (x_1, y_1) \) is \( m = 12x_1^2 - 10x_1^4 \).
The equation of the tangent line at \( (x_1, y_1) \) is:
\( y - y_1 = (12x_1^2 - 10x_1^4)(x - x_1) \)
Since the tangent passes through the origin \( (0, 0) \), we substitute \( x = 0 \) and \( y = 0 \):
\( -y_1 = (12x_1^2 - 10x_1^4)(-x_1) \)

\( \implies y_1 = 12x_1^3 - 10x_1^5 \)
Equating our two expressions for \( y_1 \):
\( 4x_1^3 - 2x_1^5 = 12x_1^3 - 10x_1^5 \)

\( \implies 8x_1^5 - 8x_1^3 = 0 \)

\( \implies 8x_1^3(x_1^2 - 1) = 0 \)
This gives the roots:
\( x_1 = 0, 1, -1 \)
Now, we find the corresponding \( y \)-coordinates from the curve equation:
- If \( x_1 = 0 \): \( y_1 = 4(0)^3 - 2(0)^5 = 0 \). Point: \( (0, 0) \).
- If \( x_1 = 1 \): \( y_1 = 4(1)^3 - 2(1)^5 = 2 \). Point: \( (1, 2) \).
- If \( x_1 = -1 \): \( y_1 = 4(-1)^3 - 2(-1)^5 = -2 \). Point: \( (-1, -2) \).
Thus, the points are \( (0, 0) \), \( (1, 2) \), and \( (-1, -2) \).
In simple words: We assume a general point on the curve, write the equation of its tangent, and force it to pass through the origin. Solving the resulting algebraic equation reveals that there are exactly three points on the curve whose tangents touch the origin.

Exam Tip: Never divide both sides of an algebraic equation by a variable (like \( x_1^3 \)) as this will lead to the loss of a valid solution (in this case, \( x_1 = 0 \)). Always factor instead.

Question 81. Find the equation of the normal to the curve x² = 4y which passes through the point (1, 2).
Answer: Let the point of contact on the curve \( x^2 = 4y \) be \( (x_1, y_1) \).
Thus, \( x_1^2 = 4y_1 \).
Differentiating the curve with respect to \( x \):
\( 2x = 4\frac{dy}{dx} \)

\( \implies \frac{dy}{dx} = \frac{x}{2} \)
The slope of the tangent at \( (x_1, y_1) \) is \( \frac{x_1}{2} \). Therefore, the slope of the normal is:
\( m_n = -\frac{2}{x_1} \)
The equation of the normal line is:
\( y - y_1 = -\frac{2}{x_1}(x - x_1) \)
Since the normal line passes through \( (1, 2) \), we substitute \( x = 1 \) and \( y = 2 \):
\( 2 - y_1 = -\frac{2}{x_1}(1 - x_1) \)

\( \implies 2 - y_1 = -\frac{2}{x_1} + 2 \)

\( \implies y_1 = \frac{2}{x_1} \)
Using the relation \( y_1 = \frac{x_1^2}{4} \):
\( \frac{x_1^2}{4} = \frac{2}{x_1} \)

\( \implies x_1^3 = 8 \)

\( \implies x_1 = 2 \)
Using this, we find \( y_1 = \frac{2^2}{4} = 1 \).
The point of contact is \( (2, 1) \), and the slope of the normal is \( m_n = -\frac{2}{2} = -1 \).
The equation of the normal is:
\( y - 1 = -1(x - 2) \)

\( \implies x + y - 3 = 0 \)
In simple words: We find the slope of the normal line at a general point, write the line equation, and substitute the given coordinates \( (1, 2) \) to solve for the exact point on the curve where the normal is drawn. This yields the line equation \( x + y - 3 = 0 \).

Exam Tip: Be sure to verify that your calculated point of contact \( (2, 1) \) actually lies on the original curve \( x^2 = 4y \) to avoid algebraic errors.

Question 82. Find the equation of the tangents at the points where the curve 2y = 3x² - 2x - 8 cuts the x-axis and show that they make supplementary angles with the x-axis.
Answer: The curve cuts the x-axis where \( y = 0 \). Setting \( y = 0 \) in the equation:
\( 3x^2 - 2x - 8 = 0 \)

\( \implies 3x^2 - 6x + 4x - 8 = 0 \)

\( \implies 3x(x - 2) + 4(x - 2) = 0 \)

\( \implies (3x + 4)(x - 2) = 0 \)
Thus, the curve cuts the x-axis at \( A(2, 0) \) and \( B\left(-\frac{4}{3}, 0\right) \).
Now, we differentiate the curve equation to find the slope of the tangents:
\( 2\frac{dy}{dx} = 6x - 2 \)

\( \implies \frac{dy}{dx} = 3x - 1 \)
- Slope of tangent at \( A(2, 0) \):
\( m_1 = 3(2) - 1 = 5 \)
Equation of tangent at \( A \):
\( y - 0 = 5(x - 2) \)

\( \implies 5x - y - 10 = 0 \)

- Slope of tangent at \( B\left(-\frac{4}{3}, 0\right) \):
\( m_2 = 3\left(-\frac{4}{3}\right) - 1 = -5 \)
Equation of tangent at \( B \):
\( y - 0 = -5\left(x + \frac{4}{3}\right) \)

\( \implies 15x + 3y + 20 = 0 \)

Let the angles made by the tangents with the positive direction of the x-axis be \( \theta_1 \) and \( \theta_2 \). Then:
\( \tan\theta_1 = m_1 = 5 \) and \( \tan\theta_2 = m_2 = -5 \)
Since \( \tan\theta_2 = -\tan\theta_1 = \tan(\pi - \theta_1) \), we have:
\( \theta_2 = \pi - \theta_1 \)

\( \implies \theta_1 + \theta_2 = \pi \)
Thus, the angles are supplementary.
In simple words: The curve crosses the x-axis at two points. Finding the slopes of the tangents at these points gives \( 5 \) and \( -5 \). Because these values are negatives of each other, the angles they form with the flat axis must add up to 180 degrees, meaning they are supplementary.

Exam Tip: Remember that two angles are supplementary if their sum is \( 180^\circ \) (or \( \pi \) radians). Showing \( \tan\theta_2 = -\tan\theta_1 \) is the standard trigonometric proof for this.

Question 83. Find the equations of the tangent and normal to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point (x₀, y₀).
Answer: The equation of the hyperbola is:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
Differentiating both sides with respect to \( x \):
\( \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0 \)

\( \implies \frac{dy}{dx} = \frac{b^2 x}{a^2 y} \)
The slope of the tangent at \( (x_0, y_0) \) is:
\( m_t = \frac{b^2 x_0}{a^2 y_0} \)
- Equation of the tangent:
\( y - y_0 = \frac{b^2 x_0}{a^2 y_0}(x - x_0) \)
Multiplying both sides by \( \frac{y_0}{b^2} \):
\( \frac{y y_0}{b^2} - \frac{y_0^2}{b^2} = \frac{x x_0}{a^2} - \frac{x_0^2}{a^2} \)

\( \implies \frac{x x_0}{a^2} - \frac{y y_0}{b^2} = \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \)
Since the point \( (x_0, y_0) \) lies on the hyperbola, we have \( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1 \). Thus, the tangent equation is:
\( \frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1 \)

- Equation of the normal:
The slope of the normal is the negative reciprocal of the tangent's slope:
\( m_n = -\frac{a^2 y_0}{b^2 x_0} \)
The equation of the normal line is:
\( y - y_0 = -\frac{a^2 y_0}{b^2 x_0}(x - x_0) \)
This can be rewritten as:
\( \frac{y - y_0}{y_0/b^2} = -\frac{x - x_0}{x_0/a^2} \)

\( \implies \frac{a^2(x - x_0)}{x_0} + \frac{b^2(y - y_0)}{y_0} = 0 \)

\( \implies \frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2 \)
In simple words: Differentiating the hyperbola's equation gives the slope of the curve. Using this, we find the formula for both the touching line (tangent) and the perpendicular line (normal) passing through the given point.

Exam Tip: Substituting the curve's identity \( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1 \) at the end simplifies the tangent line to its standard, elegant textbook form.

Question 84. A window is in the form of a rectangle surmounted by an equilateral triangle. Given that the perimeter is 16 metres. Find the width of the window in order that the maximum amount of light may be admitted.
Answer: Let the width of the window (which is also the side of the equilateral triangle and the base of the rectangle) be \( x \) metres. Let the height of the rectangular part be \( y \) metres.
The perimeter \( P \) of the window is the sum of the three outer sides of the rectangle and the two outer sides of the equilateral triangle:
\( P = x + 2y + 2x = 3x + 2y = 16 \)
From this, we express \( y \) in terms of \( x \):
\( \implies y = 8 - \frac{3}{2}x \)
The amount of light admitted is proportional to the total area \( A \) of the window:
\( A = \text{Area of rectangle} + \text{Area of triangle} \)
\( A = x y + \frac{\sqrt{3}}{4}x^2 \)
Substituting the value of \( y \):
\( A = x\left(8 - \frac{3}{2}x\right) + \frac{\sqrt{3}}{4}x^2 \)

\( \implies A = 8x - \left(\frac{6 - \sqrt{3}}{4}\right)x^2 \)
To maximize the area, we differentiate \( A \) with respect to \( x \):
\( \frac{dA}{dx} = 8 - \left(\frac{6 - \sqrt{3}}{2}\right)x \)
Setting \( \frac{dA}{dx} = 0 \) to find the optimal width:
\( \left(\frac{6 - \sqrt{3}}{2}\right)x = 8 \)

\( \implies x = \frac{16}{6 - \sqrt{3}} \text{ metres} \)
To confirm maximum area, we take the second derivative:
\( \frac{d^2A}{dx^2} = -\left(\frac{6 - \sqrt{3}}{2}\right) < 0 \)
Since the second derivative is negative, the area is indeed maximized.
Rationalizing the denominator of \( x \):
\( x = \frac{16(6 + \sqrt{3})}{36 - 3} = \frac{16(6 + \sqrt{3})}{33} \text{ metres} \)
In simple words: We find a formula for the area using only one variable (width) by substituting the height from the perimeter equation. Solving for the maximum area shows that the window's width should be \( \frac{16(6 + \sqrt{3})}{33} \) metres.

Exam Tip: Be careful when writing the perimeter expression. The boundary of the window does not include the horizontal line where the triangle sits on top of the rectangle.

Question 85. A jet of an enemy is flying along the curve y = x² + 2. A soldier is placed at the point (3, 2). What is the nearest distance between the soldier and the jet?
Answer: Let \( P(x, y) \) be any point on the curve \( y = x^2 + 2 \) representing the position of the jet [85]. Let the position of the soldier be \( S(3, 2) \) [85].
The distance \( D \) between \( P \) and \( S \) is:
\( D = \sqrt{(x - 3)^2 + (y - 2)^2} \)
Since \( y - 2 = x^2 \), we substitute this into the distance squared function \( Z = D^2 \):
\( Z = (x - 3)^2 + (x^2)^2 = x^4 + x^2 - 6x + 9 \)
To minimize the distance, we differentiate \( Z \) with respect to \( x \):
\( \frac{dZ}{dx} = 4x^3 + 2x - 6 \)
Setting \( \frac{dZ}{dx} = 0 \):
\( 4x^3 + 2x - 6 = 0 \)

\( \implies 2x^3 + x - 3 = 0 \)
By inspection, \( x = 1 \) is a root because \( 2(1)^3 + 1 - 3 = 0 \). Factoring the equation:
\( (x - 1)(2x^2 + 2x + 3) = 0 \)
Since the quadratic term has a negative discriminant (\( 2^2 - 4(2)(3) < 0 \)), it has no real roots. Thus, \( x = 1 \) is the only real critical point.
The second derivative is:
\( \frac{d^2Z}{dx^2} = 12x^2 + 2 \)
At \( x = 1 \), \( \frac{d^2Z}{dx^2} = 14 > 0 \), confirming that \( x = 1 \) is the minimum.
The nearest point on the curve is \( (1, 1^2 + 2) = (1, 3) \).
The nearest distance is:
\( D = \sqrt{(1 - 3)^2 + (3 - 2)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{5} \text{ units} \)
In simple words: We find the distance between a general point on the path of the jet and the soldier's location. Minimizing this distance using derivatives reveals that the closest the jet gets to the soldier is exactly \( \sqrt{5} \) units.

Exam Tip: Always minimize the square of the distance rather than the distance itself to keep your differentiation simple and root-free.

Question 86. Find a point on the parabola y² = 4x which is nearest to the point (2, -8).
Answer: Any point on the parabola \( y^2 = 4x \) can be represented in parametric form as \( P(t^2, 2t) \). Let the given point be \( A(2, -8) \) [86].
The distance \( D \) between \( P \) and \( A \) is:
\( D = \sqrt{(t^2 - 2)^2 + (2t + 8)^2} \)
Let \( Z = D^2 \) be the square of the distance:
\( Z = (t^2 - 2)^2 + (2t + 8)^2 \)

\( Z = t^4 - 4t^2 + 4 + 4t^2 + 32t + 64 \)

\( Z = t^4 + 32t + 68 \)
To minimize this function, we differentiate with respect to \( t \):
\( \frac{dZ}{dt} = 4t^3 + 32 \)
Setting \( \frac{dZ}{dt} = 0 \):
\( 4t^3 + 32 = 0 \)

\( \implies t^3 = -8 \)

\( \implies t = -2 \)
The second derivative is:
\( \frac{d^2Z}{dt^2} = 12t^2 \)
At \( t = -2 \), \( \frac{d^2Z}{dt^2} = 12(-2)^2 = 48 > 0 \), indicating a local minimum.
Substituting \( t = -2 \) back to find coordinates of \( P \):
\( x = (-2)^2 = 4 \)
\( y = 2(-2) = -4 \)
Thus, the point nearest to \( (2, -8) \) is \( (4, -4) \).
In simple words: By using the parametric coordinates of the parabola, we can find a formula for the distance to the point. Minimizing this formula shows that \( (4, -4) \) is the closest point on the parabola.

Exam Tip: Parametric coordinates like \( (t^2, 2t) \) for a parabola are highly recommended as they eliminate fractions and square roots from your distance function.

Question 87. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each cover and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum.
Answer: Let the side of the square cut off from each corner (referred to as "cover") be \( x \) cm.
The dimensions of the open box formed by folding up the flaps will be:
- Length = \( 18 - 2x \) cm
- Width = \( 18 - 2x \) cm
- Height = \( x \) cm
The volume \( V \) of the box is:
\( V = (18 - 2x)^2 \cdot x = (324 - 72x + 4x^2)x = 4x^3 - 72x^2 + 324x \)
Differentiating \( V \) with respect to \( x \):
\( \frac{dV}{dx} = 12x^2 - 144x + 324 = 12(x^2 - 12x + 27) = 12(x - 3)(x - 9) \)
Setting the derivative to zero for maximization:
\( 12(x - 3)(x - 9) = 0 \)

\( \implies x = 3 \text{ or } x = 9 \)
If \( x = 9 \), the length of the box becomes \( 18 - 2(9) = 0 \), which is physically impossible. Thus, we select \( x = 3 \).
The second derivative is:
\( \frac{d^2V}{dx^2} = 24x - 144 \)
At \( x = 3 \):
\( \frac{d^2V}{dx^2} = 24(3) - 144 = -72 < 0 \)
Since the second derivative is negative, \( x = 3 \) maximizes the volume.
Thus, the side of the square to be cut off is \( 3 \) cm.
In simple words: Cutting a square of side \( x \) from each corner and folding up the sides gives a box. Expressing its volume as a function of \( x \) and finding where its derivative is zero shows that cutting squares of side \( 3 \) cm yields the largest box.

Exam Tip: Always state why you rejected the other mathematical root (like \( x = 9 \)) in your written solution to demonstrate complete physical understanding.

Question 88. A window in the form of a rectangle is surmounted by a semi circular opening. The total perimeter of the window is 30 metres. Find the dimensions of the rectangular part of the window to admit maximum light through the whole opening.
Answer: Let the width of the rectangular part be \( 2r \), which means the radius of the semicircular top is \( r \). Let the height of the rectangular part be \( y \) metres.
The perimeter of the window includes three sides of the rectangle and the curved boundary of the semicircle:
\( P = 2r + 2y + \pi r = 30 \)
From this, we express \( y \) in terms of \( r \):
\( \implies 2y = 30 - r(\pi + 2) \)

\( \implies y = 15 - \left(\frac{\pi + 2}{2}\right)r \)
The total area \( A \) of the window is:
\( A = \text{Area of rectangle} + \text{Area of semicircle} \)
\( A = 2r y + \frac{1}{2}\pi r^2 \)
Substituting the value of \( y \):
\( A = 2r \left[ 15 - \left(\frac{\pi + 2}{2}\right)r \right] + \frac{1}{2}\pi r^2 \)

\( A = 30r - r^2(\pi + 2) + \frac{1}{2}\pi r^2 = 30r - \left(\frac{\pi + 4}{2}\right)r^2 \)
Differentiating \( A \) with respect to \( r \):
\( \frac{dA}{dr} = 30 - (\pi + 4)r \)
Setting the derivative to zero:
\( 30 - (\pi + 4)r = 0 \)

\( \implies r = \frac{30}{\pi + 4} \)
Taking the second derivative:
\( \frac{d^2A}{dr^2} = -(\pi + 4) < 0 \), which indicates a maximum area.
The dimensions of the rectangular part are:
- Width = \( 2r = \frac{60}{\pi + 4} \text{ metres} \)
- Height = \( y = 15 - \left(\frac{\pi + 2}{2}\right)\left(\frac{30}{\pi + 4}\right) = \frac{15(\pi + 4) - 15(\pi + 2)}{\pi + 4} = \frac{30}{\pi + 4} \text{ metres} \)
In simple words: We find a formula for the area of the window using its radius. Maximizing this area reveals that for maximum light, the width of the rectangular base should be \( \frac{60}{\pi + 4} \) m and the height should be \( \frac{30}{\pi + 4} \) m.

Exam Tip: Choosing the width of the rectangle to be \( 2r \) (instead of \( x \)) prevents the introduction of fractional halves of radius \( x/2 \) in your area and perimeter equations.

Question 89. An open box with square base is to be made out of a given iron sheet of area 27 sq. meter, show that the maximum value of the box is 13.5 cubic metres.
Answer: Let the side of the square base of the open box be \( x \) and its height be \( y \).
Since the box has an open top, its total surface area is:
\( S = x^2 + 4xy = 27 \)
From this, we write \( y \) in terms of \( x \):
\( \implies y = \frac{27 - x^2}{4x} \)
The volume \( V \) of the box is:
\( V = x^2 y \)
Substituting the value of \( y \):
\( V = x^2 \left(\frac{27 - x^2}{4x}\right) = \frac{27x - x^3}{4} \)
Differentiating \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{27 - 3x^2}{4} \)
Setting the derivative to zero:
\( 27 - 3x^2 = 0 \)

\( \implies x^2 = 9 \)

\( \implies x = 3 \text{ units (since } x > 0) \)
The second derivative is:
\( \frac{d^2V}{dx^2} = -\frac{6x}{4} = -\frac{3x}{2} \)
At \( x = 3 \), \( \frac{d^2V}{dx^2} = -4.5 < 0 \), meaning the volume is maximized.
The maximum volume of the box is:
\( V = \frac{27(3) - 3^3}{4} = \frac{81 - 27}{4} = \frac{54}{4} = 13.5 \text{ cubic metres} \)
Hence proved.
In simple words: For an open box with a fixed surface area of 27 square metres, expressing volume as a function of the base length and finding the maximum yields a maximum volume of exactly 13.5 cubic metres.

Exam Tip: Be careful not to include a second base area term when writing the surface area equation, since the box is open at the top.

Question 90. A wire of length 28 cm is to be cut into two pieces. One of the two pieces is to be made into a square and other in to a circle. What should be the length of two pieces so that the combined area of the square and the circle is minimum?
Answer: Let the wire be cut into two pieces of length \( x \) and \( 28 - x \).
Let the piece of length \( x \) be bent to form a square:
Perimeter of square = \( x \)
Side of square = \( \frac{x}{4} \)
Area of square = \( \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \)

Let the piece of length \( 28 - x \) be bent to form a circle:
Circumference of circle = \( 2\pi r = 28 - x \)
Radius of circle = \( \frac{28 - x}{2\pi} \)
Area of circle = \( \pi r^2 = \pi \left(\frac{28 - x}{2\pi}\right)^2 = \frac{(28 - x)^2}{4\pi} \)

The combined area \( A \) is:
\( A = \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi} \)
Differentiating \( A \) with respect to \( x \):
\( \frac{dA}{dx} = \frac{2x}{16} - \frac{2(28 - x)}{4\pi} = \frac{x}{8} - \frac{28 - x}{2\pi} \)
Setting the derivative to zero to find the minimum:
\( \frac{x}{8} = \frac{28 - x}{2\pi} \)

\( \implies \pi x = 4(28 - x) \)

\( \implies \pi x = 112 - 4x \)

\( \implies (\pi + 4)x = 112 \)

\( \implies x = \frac{112}{\pi + 4} \text{ cm} \)
The second derivative is:
\( \frac{d^2A}{dx^2} = \frac{1}{8} + \frac{1}{2\pi} > 0 \), confirming that this value minimizes the combined area.
The length of the other piece is:
\( 28 - x = 28 - \frac{112}{\pi + 4} = \frac{28\pi + 112 - 112}{\pi + 4} = \frac{28\pi}{\pi + 4} \text{ cm} \)
Thus, the two pieces should be of lengths \( \frac{112}{\pi + 4} \) cm and \( \frac{28\pi}{\pi + 4} \) cm.
In simple words: To get the smallest combined area, we divide the wire into two pieces such that one piece is \( \frac{112}{\pi + 4} \) cm (for the square) and the other is \( \frac{28\pi}{\pi + 4} \) cm (for the circle).

Exam Tip: Pay special attention when taking the derivative of \( (28 - x)^2 \); the chain rule introduces a negative sign that changes the operator to subtraction.

Question 91. Show that the height of the cylinder of maximum volume which can be inscribed in a sphere of radius R is \( \frac{2R}{\sqrt{3}} \). Also find the maximum volume.
Answer: Let a cylinder of radius \( r \) and height \( h \) be inscribed in a sphere of radius \( R \).
From the right-angled triangle formed by the sphere's center, we have:
\( R^2 = r^2 + \left(\frac{h}{2}\right)^2 \)
Rearranging to find \( r^2 \) in terms of \( h \):
\( \implies r^2 = R^2 - \frac{h^2}{4} \)
The volume \( V \) of the cylinder is:
\( V = \pi r^2 h \)
Substituting the expression for \( r^2 \):
\( V = \pi \left(R^2 - \frac{h^2}{4}\right)h = \pi R^2 h - \frac{\pi}{4} h^3 \)
Differentiating \( V \) with respect to \( h \):
\( \frac{dV}{dh} = \pi R^2 - \frac{3\pi}{4} h^2 \)
Setting the derivative to zero:
\( \pi R^2 = \frac{3\pi}{4} h^2 \)

\( \implies h^2 = \frac{4R^2}{3} \)

\( \implies h = \frac{2R}{\sqrt{3}} \)
To verify this is a maximum, we find the second derivative:
\( \frac{d^2V}{dh^2} = -\frac{3\pi}{2}h \)
Since \( h > 0 \), the second derivative is negative, confirming a maximum.
The maximum volume is:
\( V_{\text{max}} = \pi \left(R^2 - \frac{R^2}{3}\right)\left(\frac{2R}{\sqrt{3}}\right) = \pi \left(\frac{2R^2}{3}\right)\left(\frac{2R}{\sqrt{3}}\right) = \frac{4\pi R^3}{3\sqrt{3}} \)
In simple words: The largest cylinder you can place inside a sphere will always have a height equal to \( \frac{2}{\sqrt{3}} \) of the sphere's radius. Its volume in this case will be \( \frac{4\pi R^3}{3\sqrt{3}} \).

Exam Tip: Be careful to express the relationship between the radius of the sphere, radius of the cylinder, and the cylinder's half-height correctly using Pythagoras' theorem.

Question 92. Show that the altitude of the right circular cone of maximum volume that can be inscribed is a sphere of radius r is \( \frac{4r}{3} \).
Answer: Let \( r \) be the radius of the sphere with center \( O \). Let a cone of radius \( R_c \) and altitude \( h \) be inscribed in this sphere.
Let \( x \) be the distance from the center of the sphere \( O \) to the base of the cone. The altitude of the cone is:
\( h = r + x \)
From the geometry of the sphere and cone base, we have:
\( R_c^2 = r^2 - x^2 \)
The volume \( V \) of the cone is:
\( V = \frac{1}{3}\pi R_c^2 h \)
Substituting the value of \( R_c^2 \):
\( V = \frac{1}{3}\pi (r^2 - x^2)(r + x) \)
Differentiating \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{\pi}{3} [ (r^2 - x^2)(1) + (r + x)(-2x) ] \)

\( \frac{dV}{dx} = \frac{\pi}{3} (r + x)(r - 3x) \)
Setting \( \frac{dV}{dx} = 0 \), we get the critical point:
\( x = \frac{r}{3} \) (since \( x = -r \) is discarded).
The second derivative is:
\( \frac{d^2V}{dx^2} = -\frac{2\pi}{3}(r + 3x) \)
At \( x = \frac{r}{3} \), the second derivative is negative, indicating a maximum.
The altitude of the cone is:
\( h = r + \frac{r}{3} = \frac{4r}{3} \)
Hence proved.
In simple words: The tallest possible cone of maximum volume that fits inside a sphere has a height equal to exactly four-thirds of the sphere's radius.

Exam Tip: Maintain consistent variable notation throughout your solution to avoid confusing the radius of the sphere \( r \) with the base radius of the cone \( R_c \).

Question 93. Prove that the surface area of solid cuboid of a square base and given volume is minimum, when it is a cube.
Answer: Let the square base of the solid cuboid have side \( x \), and let its height be \( y \).
The volume \( V \) of the cuboid is constant and is given by:
\( V = x^2 y \)
From this, we write \( y \) in terms of \( x \):
\( \implies y = \frac{V}{x^2} \)
The total surface area \( S \) of the cuboid is:
\( S = 2x^2 + 4xy \)
Substituting the value of \( y \):
\( S = 2x^2 + 4x\left(\frac{V}{x^2}\right) = 2x^2 + \frac{4V}{x} \)
Differentiating \( S \) with respect to \( x \):
\( \frac{dS}{dx} = 4x - \frac{4V}{x^2} \)
Setting the derivative to zero:
\( 4x = \frac{4V}{x^2} \)

\( \implies x^3 = V \)
Substituting \( V = x^2 y \):
\( x^3 = x^2 y \)

\( \implies x = y \)
Taking the second derivative of \( S \):
\( \frac{d^2S}{dx^2} = 4 + \frac{8V}{x^3} \)
Since \( x > 0 \) and \( V > 0 \), the second derivative is positive, confirming a minimum.
Since the base is a square of side \( x \) and the height \( y \) is also equal to \( x \), all dimensions of the cuboid are equal. Thus, the cuboid is a cube.
Hence proved.
In simple words: Among all possible boxes with a square base that hold the same volume, the shape that uses the absolute least amount of material (surface area) is a perfect cube.

Exam Tip: Ensure that you show the second derivative is positive to mathematically justify that you have found a minimum rather than a maximum.

Question 94. Show that the volume of the greatest cylinder which can be inscribed in a right circular cone of height h and semi-vertical angle α is \( \frac{4}{27} \pi h^3 \tan^2\alpha \).
Answer: Let a cylinder of radius \( r \) and height \( y \) be inscribed in a right circular cone of height \( h \), base radius \( R = h \tan\alpha \), and semi-vertical angle \( \alpha \).
Using the property of similar triangles:
\( \frac{h - y}{h} = \frac{r}{R} \)

\( \implies 1 - \frac{y}{h} = \frac{r}{h \tan\alpha} \)

\( \implies y = h - r\cot\alpha \)
The volume \( V \) of the cylinder is:
\( V = \pi r^2 y \)
Substituting the value of \( y \):
\( V = \pi r^2 (h - r\cot\alpha) = \pi (h r^2 - r^3 \cot\alpha) \)
Differentiating \( V \) with respect to \( r \):
\( \frac{dV}{dr} = \pi (2h r - 3r^2 \cot\alpha) \)
Setting the derivative to zero:
\( 2h r - 3r^2 \cot\alpha = 0 \)
Since \( r \neq 0 \):
\( \implies r = \frac{2}{3}h \tan\alpha \)
Taking the second derivative of \( V \):
\( \frac{d^2V}{dr^2} = \pi (2h - 6r \cot\alpha) \)
Substituting \( r = \frac{2}{3}h \tan\alpha \):
\( \frac{d^2V}{dr^2} = \pi \left[ 2h - 6\left(\frac{2}{3}h \tan\alpha\right)\cot\alpha \right] = -2\pi h < 0 \)
This confirms that the volume is maximized.
The maximum volume is:
\( V_{\text{max}} = \pi \left(\frac{2}{3}h \tan\alpha\right)^2 \left( h - \frac{2}{3}h \tan\alpha\cot\alpha \right) \)

\( V_{\text{max}} = \pi \left(\frac{4}{9}h^2 \tan^2\alpha\right) \left(\frac{1}{3}h\right) = \frac{4}{27}\pi h^3 \tan^2\alpha \)
Hence proved.
In simple words: Using geometry, we relate the cylinder's size to the cone's dimensions. Solving for the cylinder's radius that gives the largest volume tells us that the volume is capped at exactly \( \frac{4}{27} \pi h^3 \tan^2\alpha \).

Exam Tip: Set up the similar triangle relation carefully. It is the core link that lets you express the cylinder's height in terms of its radius.

Question 95. Show that the right triangle of maximum area that can be inscribed in a circle is an isosceles triangle.
Answer: Let a right-angled triangle be inscribed in a circle of radius \( R \). Any right-angled triangle inscribed in a circle must have the circle's diameter as its hypotenuse. Thus, the length of the hypotenuse is \( 2R \).
Let the other two sides of the right triangle be \( x \) and \( y \). By the Pythagorean theorem:
\( x^2 + y^2 = 4R^2 \)

\( \implies y = \sqrt{4R^2 - x^2} \)
The area \( A \) of the triangle is:
\( A = \frac{1}{2} x y = \frac{1}{2} x \sqrt{4R^2 - x^2} \)
Let us maximize the square of the area, \( Z = A^2 \):
\( Z = \frac{1}{4} x^2 (4R^2 - x^2) = R^2 x^2 - \frac{1}{4} x^4 \)
Differentiating \( Z \) with respect to \( x \):
\( \frac{dZ}{dx} = 2R^2 x - x^3 \)
Setting the derivative to zero:
\( 2R^2 x - x^3 = 0 \)
Since \( x \neq 0 \):
\( \implies x^2 = 2R^2 \)

\( \implies x = \sqrt{2}R \)
Substituting this back to find \( y \):
\( y = \sqrt{4R^2 - 2R^2} = \sqrt{2}R \)
Since \( x = y \), the two perpendicular sides are of equal length. This shows that the triangle of maximum area is an isosceles right-angled triangle.
Taking the second derivative of \( Z \):
\( \frac{d^2Z}{dx^2} = 2R^2 - 3x^2 \)
At \( x = \sqrt{2}R \), \( \frac{d^2Z}{dx^2} = 2R^2 - 6R^2 = -4R^2 < 0 \), which confirms a maximum area.
Hence proved.
In simple words: Since any right triangle in a circle has the diameter as its longest side, we can find a formula for its area. Maximizing this area proves that both remaining sides must be equal, making it an isosceles triangle.

Exam Tip: Remember that the angle subtended by a semicircle is a right angle; this is why the hypotenuse is always the diameter.

Question 96. A given quantity of metal is to be cast half cylinder with a rectangular box and semicircular ends. Show that the total surface area is minimum when the ratio of the length of cylinder to the diameter of its semicircular ends is \( \pi : (\pi + 2) \).
Answer: Let \( r \) be the radius of the semicircular ends, and let \( l \) be the length of the half-cylinder.
The diameter of the semicircular ends is \( 2r \).
The volume \( V \) of the half-cylinder is a given constant:
\( V = \frac{1}{2}\pi r^2 l \)
From this, we express \( l \) in terms of \( r \):
\( \implies l = \frac{2V}{\pi r^2} \)
The total surface area \( S \) of this solid consists of the curved surface area, the areas of the two semicircular ends, and the area of the flat rectangular base:
\( S = \pi r l + \pi r^2 + 2r l \)
Substituting the value of \( l \):
\( S = \pi r \left(\frac{2V}{\pi r^2}\right) + \pi r^2 + 2r \left(\frac{2V}{\pi r^2}\right) \)

\( \implies S = \pi r^2 + \frac{2V}{r} + \frac{4V}{\pi r} = \pi r^2 + \frac{2V(\pi + 2)}{\pi r} \)
To minimize the surface area, we differentiate \( S \) with respect to \( r \):
\( \frac{dS}{dr} = 2\pi r - \frac{2V(\pi + 2)}{\pi r^2} \)
Setting the derivative to zero:
\( 2\pi r = \frac{2V(\pi + 2)}{\pi r^2} \)

\( \implies \pi^2 r^3 = V(\pi + 2) \)
Substitute \( V = \frac{1}{2}\pi r^2 l \) back into the equation:
\( \pi^2 r^3 = \left(\frac{1}{2}\pi r^2 l\right)(\pi + 2) \)
Dividing both sides by \( \pi r^2 \):
\( \pi r = \frac{1}{2}l(\pi + 2) \)

\( \implies 2\pi r = l(\pi + 2) \)
We want the ratio of the length of the cylinder (\( l \)) to the diameter (\( 2r \)):
\( \implies \frac{l}{2r} = \frac{\pi}{\pi + 2} \)
Thus, the ratio is \( \pi : (\pi + 2) \).
The second derivative of \( S \) is:
\( \frac{d^2S}{dr^2} = 2\pi + \frac{4V(\pi + 2)}{\pi r^3} > 0 \), confirming that this ratio minimizes the surface area.
Hence proved.
In simple words: To make a half-cylinder shape with the absolute minimum surface area for a fixed volume, the ratio of its length to its base diameter must be exactly \( \pi : (\pi + 2) \).

Exam Tip: Be sure to write the final ratio in the colon format \( \pi : (\pi + 2) \) requested by the question to complete the proof properly.

2. Water is dripping out from a conical funnel at a uniform rate of 4cm³/sec through a tiny hole at the vertex in the bottom. When the slant height of the water is 3cm, find the rate of decrease of the slant height of the water cone .Given that the vertical angle of the funnel is 120⁰.

3. Find the points on the curve y = x³- 11x + 5at which the tangent has the equation y = x- 1

4. Find the equations of the tangent and normal to the curve y= x-7/(x-2)(x-3)at the point, where it cuts x-axis.

5. Find the points on the curve 9y²= x³ where the normal to curve makes equal intercepts with the axes.