CBSE Class 12 Mathematics Indefinite and Definite Integrals Assignment Set 01

CBSE Class 12 Mathematics Indefinite and Definite Integrals Assignment Set A.Chapter wise assignments are being given by teachers to students to make them understand the chapter concepts. Its extremely critical for all CBSE students to practice all assignments which will help them in gaining better marks in examinations. All assignments available for free download on the website are developed by the best teachers having many years of teaching experience in CBSE schools all over the country. Students, teachers and parents are advised to download the CBSE study material.

Schematic Diagram

Some Important Results/Concepts

• Whenever one quantity \( y \) varies with another quantity \( x \), satisfying some rule \( y = f(x) \), then \( \frac{dy}{dx} \) (or \( f'(x) \)) represents the rate of change of \( y \) with respect to \( x \), and \( \left[ \frac{dy}{dx} \right]_{x=x_0} \) (or \( f'(x_0) \)) represents the rate of change of \( y \) with respect to \( x \) at \( x = x_0 \).
• Let \( I \) be an open interval contained in the domain of a real valued function \( f \). Then \( f \) is said to be:

(i) increasing on \( I \) if \( x_1 < x_2 \) in \( I \implies f(x_1) \le f(x_2) \) for all \( x_1, x_2 \in I \).

(ii) strictly increasing on \( I \) if \( x_1 < x_2 \) in \( I \implies f(x_1) < f(x_2) \) for all \( x_1, x_2 \in I \).

(iii) decreasing on \( I \) if \( x_1 < x_2 \) in \( I \implies f(x_1) \ge f(x_2) \) for all \( x_1, x_2 \in I \).

(iv) strictly decreasing on \( I \) if \( x_1 < x_2 \) in \( I \implies f(x_1) > f(x_2) \) for all \( x_1, x_2 \in I \).

• Derivative test for monotonicity:

(i) \( f \) is strictly increasing in \( (a, b) \) if \( f'(x) > 0 \) for each \( x \in (a, b) \).

(ii) \( f \) is strictly decreasing in \( (a, b) \) if \( f'(x) < 0 \) for each \( x \in (a, b) \).

(iii) A function will be increasing (decreasing) in \( \mathbb{R} \) if it is so in every interval of \( \mathbb{R} \).

• The slope of the tangent to the curve \( y = f(x) \) at the point \( (x_0, y_0) \) is given by \( \left[ \frac{dy}{dx} \right]_{(x_0, y_0)} = f'(x_0) \).
• The equation of the tangent at \( (x_0, y_0) \) to the curve \( y = f(x) \) is given by \( y - y_0 = f'(x_0)(x - x_0) \).
• The slope of the normal to the curve \( y = f(x) \) at \( (x_0, y_0) \) is given by \( -\frac{1}{f'(x_0)} \).
• The equation of the normal at \( (x_0, y_0) \) to the curve \( y = f(x) \) is given by \( y - y_0 = -\frac{1}{f'(x_0)}(x - x_0) \).

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• If the slope of the tangent line is zero, then \( \tan \theta = 0 \implies \theta = 0 \), which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at \( (x_0, y_0) \) is given by \( y = y_0 \).
• If \( \theta \to \frac{\pi}{2} \), then \( \tan \theta \to \infty \), which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at \( (x_0, y_0) \) is given by \( x = x_0 \).
• Increment \( \Delta y \) in the function \( y = f(x) \) corresponding to increment \( \Delta x \) in \( x \) is given by \( \Delta y \approx \frac{dy}{dx} \Delta x \).
• Relative error in \( y = \frac{\Delta y}{y} \).
• Percentage error in \( y = \frac{\Delta y}{y} \times 100 \).
• Let \( f \) be a function defined on an interval \( I \):

(a) \( f \) is said to have a maximum value in \( I \) if there exists a point \( c \in I \) such that \( f(c) \ge f(x) \) for all \( x \in I \). The number \( f(c) \) is called the maximum value of \( f \) in \( I \) and \( c \) is called a point of maximum value of \( f \).

(b) \( f \) is said to have a minimum value in \( I \) if there exists a point \( c \in I \) such that \( f(c) \le f(x) \) for all \( x \in I \). The number \( f(c) \) is called the minimum value of \( f \) in \( I \) and \( c \) is called a point of minimum value of \( f \).

(c) \( f \) is said to have an extreme value in \( I \) if there exists a point \( c \in I \) such that \( f(c) \) is either a maximum value or a minimum value of \( f \).

• Absolute Maxima and Minima:
  Let \( f \) be a function defined on the interval \( I \) and \( c \in I \). Then:

(a) \( f(c) \) is an absolute minimum if \( f(x) \ge f(c) \) for all \( x \in I \).

(b) \( f(c) \) is an absolute maximum if \( f(x) \le f(c) \) for all \( x \in I \).

(c) \( c \in I \) is called a critical point of \( f \) if \( f'(c) = 0 \) or if \( f \) is not differentiable at \( c \).

(d) The absolute maximum or minimum value of a continuous function \( f \) on a closed interval \( [a, b] \) occurs at the endpoints \( a, b \) or at critical points of \( f \).

If \( c_1, c_2, \ldots, c_n \) are the critical points lying in \( [a, b] \), then:
Absolute maximum value of \( f = \max \{ f(a), f(c_1), f(c_2), \ldots, f(c_n), f(b) \} \)
Absolute minimum value of \( f = \min \{ f(a), f(c_1), f(c_2), \ldots, f(c_n), f(b) \} \).

• Local Maxima and Minima:

(a) A function \( f \) is said to have a local maximum at \( x = a \) if \( f(a \pm h) \le f(a) \) for sufficiently small \( h > 0 \).

(b) A function \( f \) is said to have a local minimum at \( x = a \) if \( f(a \pm h) \ge f(a) \) for sufficiently small \( h > 0 \).

• First Derivative Test: A function \( f \) has a maximum at a point \( x = a \) if:

(i) \( f'(a) = 0 \), and

(ii) \( f'(x) \) changes sign from positive to negative in the neighborhood of \( a \) (as \( x \) increases from left to right).

Conversely, \( f \) has a minimum at \( x = a \) if:

(i) \( f'(a) = 0 \), and

(ii) \( f'(x) \) changes sign from negative to positive in the neighborhood of \( a \).

If \( f'(a) = 0 \) and \( f'(x) \) does not change sign, then \( f(x) \) has neither a maximum nor a minimum at \( a \), and the point \( a \) is called a point of inflection.

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• Second Derivative Test:

(i) A function has a local maximum at \( x = a \) if \( f'(a) = 0 \) and \( f''(a) < 0 \).

(ii) A function has a local minimum at \( x = a \) if \( f'(a) = 0 \) and \( f''(a) > 0 \).

Assignments

1. Rate of change

Level I

Question 1. A balloon, which always remains spherical, has a variable diameter \( \frac{3}{2}(2x + 1) \). Find the rate of change of its volume with respect to x.
Answer: Let \( d \) be the diameter of the spherical balloon:
\( d = \frac{3}{2}(2x + 1) \).
The radius \( r \) is half of the diameter:
\( r = \frac{3}{4}(2x + 1) \).
The volume \( V \) of a sphere is given by:
\( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left[ \frac{3}{4}(2x + 1) \right]^3 \)

\( \implies V = \frac{4}{3}\pi \left( \frac{27}{64} \right) (2x + 1)^3 = \frac{9}{16}\pi (2x + 1)^3 \).
We differentiate the volume \( V \) with respect to \( x \) using the chain rule:
\( \frac{dV}{dx} = \frac{d}{dx}\left[ \frac{9}{16}\pi (2x + 1)^3 \right] \)

\( \implies \frac{dV}{dx} = \frac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot \frac{d}{dx}(2x+1) \)

\( \implies \frac{dV}{dx} = \frac{27}{16}\pi (2x + 1)^2 \cdot 2 = \frac{27}{8}\pi (2x + 1)^2 \).
Thus, the rate of change of volume with respect to \( x \) is \( \frac{27}{8}\pi (2x + 1)^2 \).
In simple words: Write the volume formula in terms of x using the radius. Then, differentiate the volume expression with respect to x using the chain rule.

Exam Tip: Be careful to use the radius, which is half of the given diameter, before substituting into the volume formula.

Question 2. The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing when the side is 8 cm long ?
Answer: Let \( s \) be the side of the square and \( A \) be its area.
Given that the rate of change of the side is:
\( \frac{ds}{dt} = 4 \text{ cm/min} \).
The area of a square is:
\( A = s^2 \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( \frac{dA}{dt} = 2s \frac{ds}{dt} \).
Substituting \( s = 8 \text{ cm} \) and \( \frac{ds}{dt} = 4 \text{ cm/min} \):
\( \frac{dA}{dt} = 2(8)(4) = 64 \text{ cm}^2\text{/min} \).
Thus, the area is increasing at a rate of 64 \( \text{cm}^2\text{/min} \).
In simple words: The rate of area change depends on the side length and how fast the side is growing. Multiply twice the side length by the side's rate of growth.

Exam Tip: Remember to include the correct units (\( \text{cm}^2\text{/min} \)) for the rate of change of the area.

Question 3. The radius of a 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 and \( C \) be its circumference.
Given:
\( \frac{dr}{dt} = 0.7 \text{ cm/sec} \).
The circumference of a circle is:
\( C = 2\pi r \).
Differentiating both sides with respect to time \( t \):
\( \frac{dC}{dt} = 2\pi \frac{dr}{dt} \).
Substituting the value of \( \frac{dr}{dt} \):
\( \frac{dC}{dt} = 2\pi (0.7) = 1.4\pi \text{ cm/sec} \).
Thus, the rate of increase of the circumference is \( 1.4\pi \text{ cm/sec} \).
In simple words: Differentiating the circumference formula shows that its growth rate is simply \( 2\pi \) times the radius growth rate, which is independent of the radius itself.

Exam Tip: Note that the rate of increase of the circumference is constant and does not depend on the size of the radius.

Level II

Question 1. Find the point on the curve \( y^2 = 8x \) for which the abscissa and ordinate change at the same rate?
Answer: Let the coordinates of the point be \( (x, y) \). The abscissa is \( x \) and the ordinate is \( y \).
The rate of change of abscissa is \( \frac{dx}{dt} \), and the rate of change of ordinate is \( \frac{dy}{dt} \).
We are given that they change at the same rate:
\( \frac{dx}{dt} = \frac{dy}{dt} \).
The equation of the curve is:
\( y^2 = 8x \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( 2y \frac{dy}{dt} = 8 \frac{dx}{dt} \).
Since \( \frac{dy}{dt} = \frac{dx}{dt} \) (and assuming the rate of change is non-zero):
\( 2y \frac{dy}{dt} = 8 \frac{dy}{dt} \implies 2y = 8 \implies y = 4 \).
Substitute \( y = 4 \) back into the curve equation \( y^2 = 8x \) to find \( x \):
\( (4)^2 = 8x \implies 16 = 8x \implies x = 2 \).
Thus, the required point on the curve is \( (2, 4) \).
In simple words: Differentiate the curve's equation with respect to time. Since the rate of change of x and y are equal, substitute one for the other to find the coordinates of the point.

Exam Tip: "Abscissa" refers to the x-coordinate, and "ordinate" refers to the y-coordinate. Clearly write down this vocabulary translation first.

Question 2. A man 2 metre high walks at a uniform speed of 6km /h away from a lamp post 6 metre high. Find the rate at which the length of his shadow increases. Also find the rate at which the tip of the shadow is moving away from the lamp post.
Answer: Let \( L \) be the lamp post of height 6 m, and \( M \) be the man of height 2 m walking away from the lamp post.
Let \( x \) be the distance of the man from the lamp post, and \( y \) be the length of his shadow. The uniform walking speed is:
\( \frac{dx}{dt} = 6 \text{ km/h} \).
By similar triangles, we have:
\( \frac{\text{Height of lamp post}}{\text{Distance of shadow tip from post}} = \frac{\text{Height of man}}{\text{Length of shadow}} \)

\( \implies \frac{6}{x + y} = \frac{2}{y} \)

\( \implies 6y = 2(x + y) \implies 4y = 2x \implies y = \frac{1}{2}x \).
Differentiating with respect to time \( t \):
\( \frac{dy}{dt} = \frac{1}{2} \frac{dx}{dt} \).
Substituting \( \frac{dx}{dt} = 6 \text{ km/h} \):
\( \frac{dy}{dt} = \frac{1}{2}(6) = 3 \text{ km/h} \).
Thus, the length of his shadow increases at a rate of 3 km/h.

The tip of the shadow is at a distance \( s = x + y \) from the lamp post.
The rate of movement of the tip is:
\( \frac{ds}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = 6 + 3 = 9 \text{ km/h} \).
Thus, the tip of the shadow is moving away at a rate of 9 km/h.
In simple words: Use similar triangles to relate the man's shadow and his distance from the lamp post. Differentiating this relation gives the rate of change of shadow length and tip speed.

Exam Tip: Be sure to distinguish between the rate of shadow growth (\( dy/dt \)) and the rate of the shadow's tip movement (\( d(x+y)/dt \)).

Question 3. The length of a rectangle is increasing at the rate of 3.5 cm/sec and its breadth is decreasing at the rate of 3cm/sec. find the rate of change of the area of the rectangle when length is 12 cm and breadth is 8 cm
Answer: Let \( x \) be the length and \( y \) be the breadth of the rectangle.
Given:
Rate of change of length, \( \frac{dx}{dt} = 3.5 \text{ cm/sec} \).
Rate of change of breadth, \( \frac{dy}{dt} = -3 \text{ cm/sec} \) (negative because it is decreasing).
The area \( A \) of a rectangle is:
\( A = xy \).
Differentiating with respect to time \( t \) using the product rule:
\( \frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt} \).
Substituting \( x = 12 \text{ cm} \), \( y = 8 \text{ cm} \), \( \frac{dx}{dt} = 3.5 \), and \( \frac{dy}{dt} = -3 \):
\( \frac{dA}{dt} = 12(-3) + 8(3.5) = -36 + 28 = -8 \text{ cm}^2\text{/sec} \).
Thus, the area of the rectangle is decreasing at a rate of 8 \( \text{cm}^2\text{/sec} \).
In simple words: Since the breadth is shrinking and the length is growing, use the product rule to balance their rates. The negative result shows that the overall area is shrinking.

Exam Tip: Always use a negative sign for rates of quantities that are decreasing, such as breadth in this problem.

Level III

Question 1. A particle moves along the curve 6 y = \(x^3\) + 2., Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.
Answer: Let the position of the particle be \( (x, y) \). We are given that:
\( \frac{dy}{dt} = 8 \frac{dx}{dt} \).
The equation of the curve is:
\( 6y = x^3 + 2 \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \).
Substituting \( \frac{dy}{dt} = 8 \frac{dx}{dt} \) into the derivative equation:
\( 6 \left( 8 \frac{dx}{dt} \right) = 3x^2 \frac{dx}{dt} \)

\( \implies 48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt} \).
Since \( \frac{dx}{dt} \neq 0 \), we can divide both sides by \( 3\frac{dx}{dt} \):
\( 16 = x^2 \implies x = \pm 4 \).

We find the corresponding \( y \)-coordinates for both values of \( x \):
Case 1: If \( x = 4 \):
\( 6y = (4)^3 + 2 \implies 6y = 64 + 2 \implies 6y = 66 \implies y = 11 \).
So, the first point is \( (4, 11) \).

Case 2: If \( x = -4 \):
\( 6y = (-4)^3 + 2 \implies 6y = -64 + 2 \implies 6y = -62 \implies y = -\frac{31}{3} \).
So, the second point is \( \left(-4, -\frac{31}{3}\right) \).
Thus, the required points on the curve are \( (4, 11) \) and \( \left(-4, -\frac{31}{3}\right) \).
In simple words: Differentiate the curve equation with respect to time, substitute the relation between the rates of change of y and x, solve for x, and find the corresponding y values on the curve.

Exam Tip: Remember to write both the positive and negative solutions for \( x \) when solving \( x^2 = 16 \).

Question 2. Water is leaking from a conical funnel at the rate of 5 \(cm^3\)/sec. If the radius of the base of the funnel is 10 cm and altitude is 20 cm, Find the rate at which water level is dropping when it is 5 cm from top.
Answer: Let \( R = 10 \text{ cm} \) be the base radius and \( H = 20 \text{ cm} \) be the total height of the cone. The rate of change of volume is:
\( \frac{dV}{dt} = -5 \text{ cm}^3\text{/sec} \) (negative because water is leaking).
Let \( r \) and \( h \) be the radius and height of the water in the funnel at any instant. By similar triangles:
\( \frac{r}{h} = \frac{R}{H} = \frac{10}{20} = \frac{1}{2} \implies r = \frac{h}{2} \).
The volume \( V \) of water in the cone is:
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12} h^3 \).
Differentiating both sides with respect to time \( t \):
\( \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt} \).
We want to find \( \frac{dh}{dt} \) when the water level is 5 cm from the top, which means the height of the water from the bottom vertex is:
\( h = 20 - 5 = 15 \text{ cm} \).
Substituting \( h = 15 \) and \( \frac{dV}{dt} = -5 \):
\( -5 = \frac{\pi}{4} (15)^2 \frac{dh}{dt} \)

\( \implies -5 = \frac{225\pi}{4} \frac{dh}{dt} \)

\( \implies \frac{dh}{dt} = -\frac{20}{225\pi} = -\frac{4}{45\pi} \text{ cm/sec} \).
Thus, 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 water level using similar triangles. Find the volume in terms of height, differentiate, and solve for the rate of height change.

Exam Tip: "5 cm from the top" means the actual height of the water column from the bottom of the cone is 15 cm. Do not use 5 cm for \( h \).

Question 3. From a cylinder drum containing petrol and kept vertical, the petrol is leaking at the rate of 10 ml/sec. If the radius of the drum is 10cm and height 50cm, find the rate at which the level of the petrol is changing when petrol level is 20 cm
Answer: Let \( r = 10 \text{ cm} \) be the radius of the cylindrical drum and \( h \) be the height of the petrol level. Since the drum is cylindrical, the radius \( r \) remains constant as the petrol level changes.
Given that the rate of change of volume is:
\( \frac{dV}{dt} = -10 \text{ ml/sec} = -10 \text{ cm}^3\text{/sec} \) (since \( 1 \text{ ml} = 1 \text{ cm}^3 \)).
The volume \( V \) of a cylinder is:
\( V = \pi r^2 h \).
Since \( r = 10 \text{ cm} \) is constant:
\( V = \pi (10)^2 h = 100\pi h \).
Differentiating with respect to time \( t \):
\( \frac{dV}{dt} = 100\pi \frac{dh}{dt} \).
Substituting \( \frac{dV}{dt} = -10 \):
\( -10 = 100\pi \frac{dh}{dt} \)

\( \implies \frac{dh}{dt} = -\frac{10}{100\pi} = -\frac{1}{10\pi} \text{ cm/sec} \).
Thus, the petrol level is changing (decreasing) at a rate of \( \frac{1}{10\pi} \text{ cm/sec} \).
In simple words: For a cylinder, the radius does not change as the fluid drains. The volume is directly proportional to the height, so the rate of height change is constant.

Exam Tip: Notice that the height of the petrol level (20 cm) is extra information and does not affect the constant rate of change in a cylinder.

2. Increasing & decreasing functions

Level I

Question 1. Show that f(x) = \(x^3\) - \(6x^2\) + 18x + 5 is an increasing function for all x ∈ R.
Answer: We differentiate the function to find its derivative:
\( f'(x) = 3x^2 - 12x + 18 \).
Factoring out 3 and completing the square:
\( f'(x) = 3(x^2 - 4x + 6) \)

\( \implies f'(x) = 3[(x^2 - 4x + 4) + 2] \)

\( \implies f'(x) = 3[(x - 2)^2 + 2] \).
Since the square of any real number is always non-negative, \( (x-2)^2 \ge 0 \).
Thus, \( (x-2)^2 + 2 \ge 2 > 0 \), which implies:
\( f'(x) = 3[(x-2)^2 + 2] > 0 \) for all \( x \in \mathbb{R} \).
Since \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), the function \( f(x) \) is strictly increasing on \( \mathbb{R} \).
In simple words: Find the derivative and complete the square. Since a squared term is always positive, adding 2 ensures the slope is always positive, proving the function is increasing everywhere.

Exam Tip: Rewriting quadratic expressions in vertex form (completing the square) is the easiest way to prove they are always positive.

Question 2. Show that the function \(x^2\) - x + 1 is neither increasing nor decreasing on (0,1)
Answer: Let \( f(x) = x^2 - x + 1 \). We find its derivative:
\( f'(x) = 2x - 1 \).
We analyze the sign of \( f'(x) \) on the interval \( (0, 1) \):
Set \( f'(x) = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2} \).
This critical point divides the interval \( (0, 1) \) into two sub-intervals:
1) On \( (0, 1/2) \): For any \( x \) in this range, \( 2x < 1 \implies f'(x) < 0 \), meaning the function is decreasing.
2) On \( (1/2, 1) \): For any \( x \) in this range, \( 2x > 1 \implies f'(x) > 0 \), meaning the function is increasing.
Since \( f(x) \) decreases on one part of the interval \( (0, 1) \) and increases on the other, the function is neither increasing nor decreasing on the entire interval \( (0, 1) \).
In simple words: The slope changes sign halfway through the interval. Because the function goes down first and then goes up, it is neither purely increasing nor decreasing on the whole range.

Exam Tip: If the derivative of a function changes sign within a given interval, the function is neither increasing nor decreasing on that interval.

Question 3. Find the intervals in which the function f(x) = sin x – cos x, 0< x< 2π is increasing or decreasing.
Answer: We differentiate \( f(x) = \sin x - \cos x \):
\( f'(x) = \cos x + \sin x \).
To find the critical points, set \( f'(x) = 0 \):
\( \cos x + \sin x = 0 \implies \sin x = -\cos x \implies \tan x = -1 \).
In the interval \( 0 < x < 2\pi \), the solutions are:
\( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \).
These points divide the interval \( (0, 2\pi) \) into three sub-intervals:

1) On \( \left(0, \frac{3\pi}{4}\right) \): For example \( x = \frac{\pi}{2} \), \( f'\left(\frac{\pi}{2}\right) = \cos \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1 > 0 \). Thus, \( f(x) \) is strictly increasing.
2) On \( \left(\frac{3\pi}{4}, \frac{7\pi}{4}\right) \): For example \( x = \pi \), \( f'(\pi) = \cos \pi + \sin \pi = -1 + 0 = -1 < 0 \). Thus, \( f(x) \) is strictly decreasing.
3) On \( \left(\frac{7\pi}{4}, 2\pi\right) \): For example \( x = \frac{11\pi}{6} \), \( f'(x) > 0 \). Thus, \( f(x) \) is strictly increasing.
Thus, the function is strictly increasing in \( \left(0, \frac{3\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right) \), and strictly decreasing in \( \left(\frac{3\pi}{4}, \frac{7\pi}{4}\right) \).
In simple words: Find where the derivative is zero to identify the boundary points. Test values in each region to determine where the function rises and where it falls.

Exam Tip: Test simple quadrant angles (like \( \pi/2, \pi \)) in your sub-intervals to quickly determine the sign of the derivative.

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Level II

Question 1. Indicate the interval in which the function f(x) = cos x, 0≤ x ≤ 2π is decreasing.
Answer: We differentiate \( f(x) = \cos x \):
\( f'(x) = -\sin x \).
A function is decreasing where its derivative is non-positive:
\( f'(x) \le 0 \implies -\sin x \le 0 \implies \sin x \ge 0 \).
In the interval \( [0, 2\pi] \), the sine function is non-negative in the first and second quadrants, i.e., when:
\( x \in [0, \pi] \).
Thus, the function \( f(x) = \cos x \) is decreasing on the interval \( [0, \pi] \).
In simple words: The derivative is negative sine. Since sine is positive from 0 to 180 degrees, the derivative is negative, meaning the cosine curve falls in that region.

Exam Tip: Remember that "decreasing" allows for the boundary endpoints where the derivative is zero, so use closed brackets \( [0, \pi] \).

Question 2. Show that the function f(x) = \( \frac{\sin x}{x} \) is strictly decreasing on ( 0, π/2)
Answer: We differentiate \( f(x) = \frac{\sin x}{x} \) using the quotient rule:
\( f'(x) = \frac{x\cos x - \sin x}{x^2} = \frac{\cos x(x - \tan x)}{x^2} \).
We analyze the signs of the terms on the interval \( (0, \pi/2) \):
1) \( x^2 > 0 \) is always positive.
2) \( \cos x > 0 \) is positive in the first quadrant.
3) For any \( x \in (0, \pi/2) \), we have the standard inequality \( \tan x > x \implies x - \tan x < 0 \) (negative).
Thus, the numerator is the product of a positive and a negative term, making it negative:
\( f'(x) = \frac{\cos x(x - \tan x)}{x^2} < 0 \) for all \( x \in (0, \pi/2) \).
Since \( f'(x) < 0 \), the function is strictly decreasing on \( (0, \pi/2) \).
In simple words: Differentiate using the quotient rule. In the first quadrant, tangent is always larger than the angle itself, making the numerator negative, which proves the function is decreasing.

Exam Tip: The inequality \( \tan x > x \) for \( x \in (0, \pi/2) \) is a standard trigonometric fact that is very useful in calculus proofs.

Question 3. Find the intervals in which the function f(x) = \( \frac{\log x}{x} \) increasing or decreasing.
Answer: The domain of \( f(x) = \frac{\log x}{x} \) is \( (0, \infty) \) because the logarithm is only defined for positive numbers.
We find the derivative using the quotient rule:
\( f'(x) = \frac{x\left(\frac{1}{x}\right) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2} \).
To find critical points, set \( f'(x) = 0 \):
\( 1 - \log x = 0 \implies \log x = 1 \implies x = e \).
This critical point divides the domain into two intervals:

1) On \( (0, e) \): For \( x < e \), \( \log x < 1 \implies 1 - \log x > 0 \). Since \( f'(x) > 0 \), the function is strictly increasing.
2) On \( (e, \infty) \): For \( x > e \), \( \log x > 1 \implies 1 - \log x < 0 \). Since \( f'(x) < 0 \), the function is strictly decreasing.
Thus, the function is strictly increasing in \( (0, e) \) and strictly decreasing in \( (e, \infty) \).
In simple words: Differentiate and find where the derivative is zero, which happens at x = e. The function rises before this point and falls after it.

Exam Tip: Be sure to explicitly state the domain limitation \( x > 0 \) at the start of any logarithmic analysis.

Level III

Question 1. Find the interval of monotonocity of the function f(x) = \(2x^2\) – log x , x ≠ 0
Answer: Since \( \log x \) is only defined for positive real numbers, the actual domain of the function is \( (0, \infty) \).
We differentiate the function:
\( f'(x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x} \).
To find the critical points, set \( f'(x) = 0 \):
\( 4x^2 - 1 = 0 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \).
Since \( x > 0 \), we only consider the critical point \( x = \frac{1}{2} \).
This point divides our domain \( (0, \infty) \) into two intervals:

1) On \( \left(0, \frac{1}{2}\right) \): For \( x < 1/2 \), \( 4x^2 < 1 \implies 4x^2 - 1 < 0 \). Since the denominator \( x > 0 \), we have \( f'(x) < 0 \). Thus, \( f(x) \) is strictly decreasing.
2) On \( \left(\frac{1}{2}, \infty\right) \): For \( x > 1/2 \), \( 4x^2 > 1 \implies 4x^2 - 1 > 0 \). Thus, \( f'(x) > 0 \). Hence, \( f(x) \) is strictly increasing.
Therefore, the function is strictly decreasing in \( \left(0, \frac{1}{2}\right) \) and strictly increasing in \( \left(\frac{1}{2}, \infty\right) \).
In simple words: Find the derivative and solve for when it is zero (at x = 1/2). The function falls from 0 to 1/2 and rises from 1/2 onward.

Exam Tip: Always discard negative critical points (like \( x = -1/2 \)) if they lie outside the natural domain of the logarithmic function.

Question 2. Prove that the function y = \( \frac{4\sin \theta}{2 + \cos \theta} - \theta \) is an increasing function of  in [ 0, π/2]
Answer: We differentiate \( y \) with respect to \( \theta \) using the quotient rule:
\( \frac{dy}{d\theta} = \frac{(2+\cos \theta)(4\cos \theta) - (4\sin \theta)(-\sin \theta)}{(2+\cos \theta)^2} - 1 \)

\( \implies \frac{dy}{d\theta} = \frac{8\cos \theta + 4\cos^2 \theta + 4\sin^2 \theta}{(2+\cos \theta)^2} - 1 \)

\( \implies \frac{dy}{d\theta} = \frac{8\cos \theta + 4}{(2+\cos \theta)^2} - 1 \)
Combining these terms over a common denominator:
\( \frac{dy}{d\theta} = \frac{8\cos \theta + 4 - (4 + 4\cos \theta + \cos^2 \theta)}{(2+\cos \theta)^2} = \frac{4\cos \theta - \cos^2 \theta}{(2+\cos \theta)^2} = \frac{\cos \theta(4 - \cos \theta)}{(2+\cos \theta)^2} \).
We analyze the signs of the terms on the interval \( [0, \pi/2] \):
1) \( (2+\cos \theta)^2 > 0 \) is always positive.
2) Since \( -1 \le \cos \theta \le 1 \), we have \( 4 - \cos \theta \ge 3 > 0 \) (always positive).
3) In the first quadrant \( [0, \pi/2] \), \( \cos \theta \ge 0 \).
Thus, the derivative is non-negative:
\( \frac{dy}{d\theta} \ge 0 \) for all \( \theta \in [0, \pi/2] \).
This proves that the function is an increasing function of \( \theta \) on \( [0, \pi/2] \).
In simple words: Differentiate using the quotient rule and simplify. Since all terms in the simplified derivative are positive in the first quadrant, the function is increasing.

Exam Tip: Use the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to simplify the numerator before combining fractions.

3. Tangents & Normals

Level I

Question 1. Find the equations of the normals to the curve \(3x^2 - y^2 = 8\) which are parallel to the line x + 3y = 4.
Answer: Let the point of contact be \( (x_1, y_1) \) on the curve.
Differentiating the curve equation implicitly with respect to \( x \):
\( 6x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x}{y} \).
The slope of the tangent at \( (x_1, y_1) \) is \( \frac{3x_1}{y_1} \). Thus, the slope of the normal is:
\( m_n = -\frac{y_1}{3x_1} \).

The given line is \( x + 3y = 4 \implies 3y = -x + 4 \implies y = -\frac{1}{3}x + \frac{4}{3} \).
Its slope is \( m = -\frac{1}{3} \). Since the normal is parallel to this line:
\( -\frac{y_1}{3x_1} = -\frac{1}{3} \implies y_1 = x_1 \).

Since the point \( (x_1, y_1) \) lies on the curve \( 3x^2 - y^2 = 8 \):
\( 3x_1^2 - x_1^2 = 8 \implies 2x_1^2 = 8 \implies x_1^2 = 4 \implies x_1 = \pm 2 \).
If \( x_1 = 2 \), then \( y_1 = 2 \), giving point \( (2, 2) \).
If \( x_1 = -2 \), then \( y_1 = -2 \), giving point \( (-2, -2) \).

Now, we write the equations of the normals with slope \( -\frac{1}{3} \):
1) At \( (2, 2) \): \( y - 2 = -\frac{1}{3}(x - 2) \implies 3y - 6 = -x + 2 \implies x + 3y = 8 \).
2) At \( (-2, -2) \): \( y + 2 = -\frac{1}{3}(x + 2) \implies 3y + 6 = -x - 2 \implies x + 3y = -8 \).
Thus, the equations of the normals are \( x + 3y = 8 \) and \( x + 3y = -8 \).
In simple words: Find the slope of the normal using the derivative. Set it equal to the slope of the given line to find the points on the curve, then write the line equations.

Exam Tip: Remember that parallel lines have equal slopes: \( m_{\text{normal}} = m_{\text{line}} \).

Question 2. Find the point on the curve \(y = x^2\) where the slope of the tangent is equal to the x-coordinate of the point.
Answer: Let the required point be \( (x_0, y_0) \).
Differentiating the curve equation \( y = x^2 \):
\( \frac{dy}{dx} = 2x \).
The slope of the tangent at \( (x_0, y_0) \) is:
\( m = 2x_0 \).
We are given that this slope equals the x-coordinate of the point:
\( 2x_0 = x_0 \implies x_0 = 0 \).
Substituting \( x_0 = 0 \) back into the curve equation to find \( y_0 \):
\( y_0 = (0)^2 = 0 \).
Thus, the required point on the curve is \( (0, 0) \).
In simple words: Find the derivative of the curve, set it equal to the x-coordinate, solve for x, and plug it back into the curve equation to get the point.

Exam Tip: Don't forget that the final answer must be a coordinate pair \( (x, y) \), not just the value of \( x \).

Question 3. At what points on the circle \(x^2 + y^2 – 2x – 4y + 1 = 0\), the tangent is parallel to x axis ?
Answer: If the tangent is parallel to the x-axis, its slope must be zero:
\( \frac{dy}{dx} = 0 \).
Differentiating the circle equation implicitly with respect to \( x \):
\( 2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} = 0 \)

\( \implies \frac{dy}{dx} (2y - 4) = 2 - 2x \)

\( \implies \frac{dy}{dx} = \frac{2 - 2x}{2y - 4} = \frac{1-x}{y-2} \).
Setting the derivative to zero:
\( \frac{1-x}{y-2} = 0 \implies 1 - x = 0 \implies x = 1 \).
Now, substitute \( x = 1 \) back into the circle equation to find \( y \):
\( (1)^2 + y^2 - 2(1) - 4y + 1 = 0 \)

\( \implies 1 + y^2 - 2 - 4y + 1 = 0 \)

\( \implies y^2 - 4y = 0 \implies y(y-4) = 0 \implies y = 0 \) or \( y = 4 \).
Thus, the points are \( (1, 0) \) and \( (1, 4) \).
In simple words: Set the derivative of the circle equal to zero to find that x = 1. Plug x = 1 back into the circle's equation to find the two matching y-coordinates.

Exam Tip: A fraction is equal to zero only when its numerator is zero and its denominator is non-zero. Check that \( y \neq 2 \) at your final points.

Level II

Question 1. Find the equation of the normal to the curve \(ay^2 = x^3\) at the point ( \(am^2\), \(am^3\))
Answer: Differentiating the curve equation implicitly with respect to \( x \):
\( 2ay \frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{2ay} \).
Evaluating the slope of the tangent at the point \( (am^2, am^3) \):
\( m_t = \frac{3(am^2)^2}{2a(am^3)} = \frac{3a^2 m^4}{2a^2 m^3} = \frac{3}{2}m \).
The slope of the normal is the negative reciprocal of the tangent slope:
\( m_n = -\frac{2}{3m} \).

Now, we write the equation of the normal line:
\( y - am^3 = -\frac{2}{3m}(x - am^2) \)

\( \implies 3m(y - am^3) = -2(x - am^2) \)

\( \implies 3my - 3am^4 = -2x + 2am^2 \)

\( \implies 2x + 3my - am^2(2 + 3m^2) = 0 \).
This is the required equation of the normal.
In simple words: Find the derivative to get the tangent slope, find the negative reciprocal to get the normal slope, and use the point-slope formula to write the line equation.

Exam Tip: Be careful with algebraic simplification; group the constants together at the end to present the line equation in standard form.

Question 2. For the curve \(y = 2x^2 + 3x + 18\), find all the points at which the tangent passes through the origin.
Answer: Let the point of contact on the curve be \( (x_1, y_1) \). This point satisfies the curve equation:
\( y_1 = 2x_1^2 + 3x_1 + 18 \).
Differentiating the curve equation:
\( \frac{dy}{dx} = 4x + 3 \).
The slope of the tangent at \( (x_1, y_1) \) is \( 4x_1 + 3 \).
The equation of the tangent at \( (x_1, y_1) \) is:
\( y - y_1 = (4x_1 + 3)(x - x_1) \).
Since this tangent passes through the origin \( (0, 0) \), we substitute \( x = 0 \) and \( y = 0 \):
\( 0 - y_1 = (4x_1 + 3)(0 - x_1) \)

\( \implies -y_1 = -x_1(4x_1 + 3) \implies y_1 = 4x_1^2 + 3x_1 \).
Equating the two expressions for \( y_1 \):
\( 2x_1^2 + 3x_1 + 18 = 4x_1^2 + 3x_1 \)

\( \implies 2x_1^2 = 18 \implies x_1^2 = 9 \implies x_1 = \pm 3 \).
Now, find the corresponding \( y \)-coordinates:
If \( x_1 = 3 \): \( y_1 = 4(3)^2 + 3(3) = 36 + 9 = 45 \). Point is \( (3, 45) \).
If \( x_1 = -3 \): \( y_1 = 4(-3)^2 + 3(-3) = 36 - 9 = 27 \). Point is \( (-3, 27) \).
Thus, the points are \( (3, 45) \) and \( (-3, 27) \).
In simple words: Write the equation of the tangent at a general point, substitute the origin (0,0) into it, and solve the resulting equation to find the coordinates.

Exam Tip: Endpoints or transition coordinates are found by solving the intersection of the tangent line equation and the curve equation.

Question 3. Find the equation of the normals to the curve \(y = x^3 + 2x + 6\) which are parallel to the line x + 14y + 4= 0
Answer: Let the point of contact be \( (x_1, y_1) \).
Differentiating the curve equation:
\( \frac{dy}{dx} = 3x^2 + 2 \).
The slope of the tangent is \( 3x_1^2 + 2 \), so the slope of the normal is:
\( m_n = -\frac{1}{3x_1^2 + 2} \).
The given line is \( x + 14y + 4 = 0 \implies 14y = -x - 4 \implies y = -\frac{1}{14}x - \frac{4}{14} \).
Its slope is \( -\frac{1}{14} \). Since the normal is parallel to this line:
\( -\frac{1}{3x_1^2 + 2} = -\frac{1}{14} \)

\( \implies 3x_1^2 + 2 = 14 \implies 3x_1^2 = 12 \implies x_1^2 = 4 \implies x_1 = \pm 2 \).

Find the corresponding \( y \)-coordinates using the curve equation:
If \( x_1 = 2 \): \( y_1 = 2^3 + 2(2) + 6 = 8 + 4 + 6 = 18 \). Point is \( (2, 18) \).
If \( x_1 = -2 \): \( y_1 = (-2)^3 + 2(-2) + 6 = -8 - 4 + 6 = -6 \). Point is \( (-2, -6) \).

Now, write the equations of the normals with slope \( -\frac{1}{14} \):
1) At \( (2, 18) \): \( y - 18 = -\frac{1}{14}(x - 2) \implies 14y - 252 = -x + 2 \implies x + 14y = 254 \).
2) At \( (-2, -6) \): \( y + 6 = -\frac{1}{14}(x + 2) \implies 14y + 84 = -x - 2 \implies x + 14y = -86 \).
The equations of the normals are \( x + 14y = 254 \) and \( x + 14y = -86 \).
In simple words: Set the formula for the normal's slope equal to the given line's slope, solve for the points of contact, and write the equations of the lines.

Exam Tip: Be sure to use the curve equation \( y = x^3 + 2x + 6 \) to find the y-coordinates of the points, not the line equation.

Question 4. Show that the equation of tangent at (x1 , y1) to the parabola \(yy_1 = 2a(x + x_1)\). 
Answer: The standard equation of the parabola is \( y^2 = 4ax \).
Differentiating both sides with respect to \( x \):
\( 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \).
The slope of the tangent at \( (x_1, y_1) \) is:
\( m = \frac{2a}{y_1} \).
The equation of the tangent line at \( (x_1, y_1) \) is given by point-slope form:
\( y - y_1 = \frac{2a}{y_1}(x - x_1) \)

\( \implies y_1(y - y_1) = 2a(x - x_1) \)

\( \implies yy_1 - y_1^2 = 2ax - 2ax_1 \).
Since the point \( (x_1, y_1) \) lies on the parabola \( y^2 = 4ax \), we have \( y_1^2 = 4ax_1 \). Substituting this in:
\( yy_1 - 4ax_1 = 2ax - 2ax_1 \)

\( \implies yy_1 = 2ax + 2ax_1 \)

\( \implies yy_1 = 2a(x + x_1) \).
Hence, proved.
In simple words: Differentiate the parabola's equation to find the slope of the tangent, write the line equation, and use the fact that the point lies on the curve to simplify the terms.

Exam Tip: Substituting the curve equation property \( y_1^2 = 4ax_1 \) is the key step to getting the final simplified equation.

Level III

Question 1. Find the equation of the tangent line to the curve y = \( \sqrt{5x-3} \) -2 which is parallel to the line 4x -2y +3 =0
Answer: Let the point of contact be \( (x_1, y_1) \).
Differentiating the curve equation \( y = \sqrt{5x-3} - 2 \):
\( \frac{dy}{dx} = \frac{5}{2\sqrt{5x-3}} \).
The given line is \( 4x - 2y + 3 = 0 \implies 2y = 4x + 3 \implies y = 2x + \frac{3}{2} \).
Its slope is \( m = 2 \). Since the tangent is parallel to this line:
\( \frac{5}{2\sqrt{5x_1-3}} = 2 \)

\( \implies 4\sqrt{5x_1-3} = 5 \)

\( \implies \sqrt{5x_1-3} = \frac{5}{4} \).
Squaring both sides:
\( 5x_1 - 3 = \frac{25}{16} \)

\( \implies 5x_1 = \frac{25}{16} + 3 = \frac{73}{16} \implies x_1 = \frac{73}{80} \).
Now, find the corresponding \( y_1 \) using the curve equation:
\( y_1 = \sqrt{5\left(\frac{73}{80}\right)-3} - 2 = \sqrt{\frac{73}{16}-3} - 2 = \sqrt{\frac{25}{16}} - 2 = \frac{5}{4} - 2 = -\frac{3}{4} \).
So, the point of contact is \( \left(\frac{73}{80}, -\frac{3}{4}\right) \).

Now, write the equation of the tangent with slope 2:
\( y + \frac{3}{4} = 2\left(x - \frac{73}{80}\right) \)

\( \implies y + \frac{3}{4} = 2x - \frac{73}{40} \)

\( \implies 2x - y = \frac{3}{4} + \frac{73}{40} = \frac{30 + 73}{40} = \frac{103}{40} \)

\( \implies 80x - 40y - 103 = 0 \).
This is the required equation of the tangent.
In simple words: Find the derivative to get the tangent slope, set it equal to 2, solve for the point of contact, and use the point-slope formula to write the tangent line equation.

Exam Tip: Take extra care with fraction arithmetic when squaring and solving for the point coordinates.

Question 2. Show that the curve \(x^2\) + \(y^2\) -2x = 0 and \(x^2\) + \(y^2\) -2y = 0 cut orthogonally at the point (0,0)
Answer: Two curves cut orthogonally if the product of their tangent slopes at the intersection point is \( -1 \), i.e., \( m_1 m_2 = -1 \).
Let the first curve be \( C_1 \): \( x^2 + y^2 - 2x = 0 \).
Differentiating implicitly:
\( 2x + 2y \frac{dy}{dx} - 2 = 0 \implies \frac{dy}{dx} = \frac{2-2x}{2y} = \frac{1-x}{y} \).
So, the slope of the tangent \( m_1 \) at \( (x, y) \) is \( \frac{1-x}{y} \).

Let the second curve be \( C_2 \): \( x^2 + y^2 - 2y = 0 \).
Differentiating implicitly:
\( 2x + 2y \frac{dy}{dx} - 2\frac{dy}{dx} = 0 \implies \frac{dy}{dx}(2y-2) = -2x \implies \frac{dy}{dx} = -\frac{x}{y-1} \).
So, the slope of the tangent \( m_2 \) at \( (x, y) \) is \( -\frac{x}{y-1} \).

At the intersection point \( (0, 0) \), let's find the limits of the slopes:
For \( C_1 \), as we approach \( (0,0) \), the tangent line is vertical since \( \frac{1-x}{y} \to \infty \) (parallel to the y-axis, equation \( x = 0 \)).
For \( C_2 \), at \( (0, 0) \), the slope \( m_2 = -\frac{0}{0-1} = 0 \), which means the tangent line is horizontal (parallel to the x-axis, equation \( y = 0 \)).
Since one tangent line is horizontal (slope 0) and the other is vertical (slope undefined/\( \infty \)), they are perpendicular to each other. Thus, the curves intersect orthogonally at \( (0, 0) \).
In simple words: Differentiate both curves. At the origin, one curve has a horizontal tangent line and the other has a vertical tangent line. Since horizontal and vertical lines are perpendicular, the curves cut orthogonally.

Exam Tip: When one slope is 0 and the other is undefined, show that the lines are parallel to the coordinate axes to prove they are perpendicular.

Page 37

Question 3. Find the condition for the curves \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and xy = \(c^2\) to intersect orthogonally.
Answer: Let the curves intersect at \( (x_1, y_1) \).
Differentiating the first curve \( C_1 \): \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \):
\( \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{b^2 x}{a^2 y} \).
So, the slope of the tangent \( m_1 \) at \( (x_1, y_1) \) is:
\( m_1 = \frac{b^2 x_1}{a^2 y_1} \).

Differentiating the second curve \( C_2 \): \( xy = c^2 \):
\( y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \).
So, the slope of the tangent \( m_2 \) at \( (x_1, y_1) \) is:
\( m_2 = -\frac{y_1}{x_1} \).

For orthogonal intersection, the product of the slopes must be \( -1 \):
\( m_1 m_2 = -1 \)

\( \implies \left( \frac{b^2 x_1}{a^2 y_1} \right) \left( -\frac{y_1}{x_1} \right) = -1 \)

\( \implies -\frac{b^2}{a^2} = -1 \)

\( \implies a^2 = b^2 \).
Thus, the condition for orthogonal intersection is \( a^2 = b^2 \).
In simple words: Find the slopes of both curves at their intersection point. Multiplying their slopes and setting the product to -1 shows that the constants a-squared and b-squared must be equal.

Exam Tip: Notice how the coordinates of the intersection point cancel out, meaning the orthogonality condition depends solely on the constants \( a \) and \( b \).

4. Approximations

Level I

Question 1. Evaluate \( \sqrt{25.3} \)
Answer: Let \( y = f(x) = \sqrt{x} \). We choose a perfect square near 25.3:
\( x = 25 \quad \text{and} \quad \Delta x = 0.3 \).
Differentiating \( f(x) \):
\( f'(x) = \frac{1}{2\sqrt{x}} \).
The approximation formula is:
\( f(x + \Delta x) \approx f(x) + f'(x)\Delta x \)

\( \implies \sqrt{25.3} \approx \sqrt{25} + \frac{1}{2\sqrt{25}}(0.3) \)

\( \implies \sqrt{25.3} \approx 5 + \frac{1}{10}(0.3) = 5 + 0.03 = 5.03 \).
Thus, the approximate value of \( \sqrt{25.3} \) is 5.03.
In simple words: Use the derivative of the square root function at the nearby perfect square 25 to estimate the small increase of 0.3.

Exam Tip: Always choose the value of \( x \) such that \( f(x) \) and \( f'(x) \) are easy to calculate mentally.

Question 2. Use differentials to approximate the cube root of 66
Answer: Let \( y = f(x) = x^{1/3} \). We choose a perfect cube near 66:
\( x = 64 \quad \text{and} \quad \Delta x = 2 \).
Differentiating \( f(x) \):
\( f'(x) = \frac{1}{3x^{2/3}} \).
The approximation formula is:
\( f(x + \Delta x) \approx f(x) + f'(x)\Delta x \)

\( \implies (66)^{1/3} \approx (64)^{1/3} + \frac{1}{3(64)^{2/3}}(2) \)
Since \( (64)^{1/3} = 4 \) and \( (64)^{2/3} = 16 \):
\( (66)^{1/3} \approx 4 + \frac{2}{48} = 4 + \frac{1}{24} \approx 4 + 0.0417 = 4.0417 \).
Thus, the approximate value of the cube root of 66 is 4.042.
In simple words: Use the derivative of the cube root function at 64 to estimate the change when we increase the number by 2.

Exam Tip: Keep your decimal division precise to at least three or four decimal places for approximation questions.

Question 3. Evaluate \( \sqrt{0.082} \)
Answer: Let \( y = f(x) = \sqrt{x} \). We choose a perfect square near 0.082:
\( x = 0.09 \quad \text{and} \quad \Delta x = -0.008 \).
Differentiating \( f(x) \):
\( f'(x) = \frac{1}{2\sqrt{x}} \).
The approximation formula is:
\( f(x + \Delta x) \approx f(x) + f'(x)\Delta x \)

\( \implies \sqrt{0.082} \approx \sqrt{0.09} + \frac{1}{2\sqrt{0.09}}(-0.008) \)

\( \implies \sqrt{0.082} \approx 0.3 - \frac{0.008}{2(0.3)} = 0.3 - \frac{0.008}{0.6} = 0.3 - 0.0133 = 0.2867 \).
Thus, the approximate value of \( \sqrt{0.082} \) is 0.287.
In simple words: Since 0.09 is a perfect square (its root is 0.3), use it as the base point with a small negative step of -0.008 to find the root.

Exam Tip: Be careful with the negative sign of \( \Delta x \) when the value is smaller than the chosen perfect square.

Question 4. Evaluate \( \sqrt{49.5} \) 
Answer: Let \( y = f(x) = \sqrt{x} \). We choose a perfect square near 49.5:
\( x = 49 \quad \text{and} \quad \Delta x = 0.5 \).
Differentiating \( f(x) \):
\( f'(x) = \frac{1}{2\sqrt{x}} \).
The approximation formula is:
\( f(x + \Delta x) \approx f(x) + f'(x)\Delta x \)

\( \implies \sqrt{49.5} \approx \sqrt{49} + \frac{1}{2\sqrt{49}}(0.5) \)

\( \implies \sqrt{49.5} \approx 7 + \frac{0.5}{14} = 7 + \frac{1}{28} \approx 7 + 0.0357 = 7.0357 \).
Thus, the approximate value of \( \sqrt{49.5} \) is 7.036.
In simple words: Use the derivative of the square root function at the nearby perfect square 49 to estimate the small increase of 0.5.

Exam Tip: Clearly state the values of \( x \) and \( \Delta x \) before writing down the approximation formula.

Level II

Question 1. If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area
Answer: Let \( r \) be the radius and \( S \) be the surface area of the sphere.
Given:
Radius, \( r = 9 \text{ cm} \).
Error in radius, \( \Delta r = 0.03 \text{ cm} \).
The surface area of a sphere is:
\( S = 4\pi r^2 \).
Differentiating with respect to \( r \):
\( \frac{dS}{dr} = 8\pi r \).
The approximate error in surface area \( \Delta S \) is:
\( \Delta S \approx \frac{dS}{dr} \Delta r \)

\( \implies \Delta S = (8\pi r) \Delta r \).
Substituting \( r = 9 \) and \( \Delta r = 0.03 \):
\( \Delta S = 8\pi (9)(0.03) = 2.16\pi \text{ cm}^2 \).
Thus, the approximate error in surface area is \( 2.16\pi \text{ cm}^2 \).
In simple words: Find the derivative of the surface area formula, and multiply it by the measured error in the radius to get the error in the area.

Exam Tip: Remember to express the final error with the correct unit of area (\( \text{cm}^2 \)).

5 Maxima & Minima

Level I

Question 1. Find the maximum and minimum value of the function f(x) = 3 – 2 sin x
Answer: We use the standard range of the sine function:
\( -1 \le \sin x \le 1 \).
Multiply the entire inequality by -2 (which reverses the inequality signs):
\( 2 \ge -2\sin x \ge -2 \)

\( \implies -2 \le -2\sin x \le 2 \).
Now, add 3 to all parts of the inequality:
\( 3 - 2 \le 3 - 2\sin x \le 3 + 2 \)

\( \implies 1 \le f(x) \le 5 \).
Thus, the maximum value of the function is 5, and the minimum value of the function is 1.
In simple words: Since the sine function fluctuates between -1 and 1, the expression \( 3 - 2\sin x \) fluctuates between \( 3 - 2(1) = 1 \) and \( 3 - 2(-1) = 5 \).

Exam Tip: For simple trigonometric functions, using the known range of sine or cosine is much faster than using calculus derivatives.

Question 2. Show that the function f(x) = \(x^3\)+\(x^2\) + x + 1 has neither a maximum value nor a minimum value
Answer: To find extreme values, we first find the derivative of the function:
\( f'(x) = 3x^2 + 2x + 1 \).
We check for critical points by setting \( f'(x) = 0 \):
\( 3x^2 + 2x + 1 = 0 \).
This is a quadratic equation. Let us find its discriminant \( D \):
\( D = b^2 - 4ac = 2^2 - 4(3)(1) = 4 - 12 = -8 \).
Since the discriminant is negative (\( D < 0 \)), the equation has no real roots. Thus, \( f'(x) \) is never zero for any real \( x \).
Since there are no real critical points, the function has neither a maximum nor a minimum value. In fact, since the leading coefficient is positive, \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), meaning the function is strictly increasing everywhere.
In simple words: Since the derivative is a quadratic function with no real roots, the slope of the curve is always positive and never becomes zero, meaning there are no peaks or valleys.

Exam Tip: If the derivative of a function is always strictly positive (or strictly negative), the function cannot have any local extrema.

Question 3. Find two positive numbers whose sum is 24 and whose product is maximum
Answer: Let the two positive numbers be \( x \) and \( y \). We are given:
\( x + y = 24 \implies y = 24 - x \).
We want to maximize their product \( P \):
\( P = xy = x(24 - x) = 24x - x^2 \).
Differentiating \( P \) with respect to \( x \):
\( \frac{dP}{dx} = 24 - 2x \).
To find the critical point, set \( \frac{dP}{dx} = 0 \):
\( 24 - 2x = 0 \implies x = 12 \).
Now, find the second derivative to verify maximality:
\( \frac{d^2P}{dx^2} = -2 \).
Since the second derivative is negative (\( -2 < 0 \)), \( x = 12 \) is a point of maximum product.
The second number is:
\( y = 24 - 12 = 12 \).
Thus, the two positive numbers are 12 and 12.
In simple words: Set up the product in terms of a single variable, differentiate, and set the derivative to zero. The second derivative is negative, confirming we found the maximum product.

Exam Tip: For a given sum, the product of two numbers is always maximized when the two numbers are equal.

Level II

Question 1. Prove that the area of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Answer: Let \( h \) be the constant hypotenuse of the right-angled triangle. Let the other two sides be \( x \) and \( y \). By Pythagoras' theorem:
\( x^2 + y^2 = h^2 \implies y = \sqrt{h^2 - x^2} \).
The area \( A \) of the right-angled triangle is:
\( A = \frac{1}{2} x y = \frac{1}{2} x \sqrt{h^2 - x^2} \).
To make calculation simpler, we maximize \( A^2 \) instead of \( A \). Let \( Z = A^2 \):
\( Z = \frac{1}{4} x^2 (h^2 - x^2) = \frac{1}{4} (h^2 x^2 - x^4) \).
Differentiating \( Z \) with respect to \( x \):
\( \frac{dZ}{dx} = \frac{1}{4} (2h^2 x - 4x^3) \).
Setting \( \frac{dZ}{dx} = 0 \):
\( 2h^2 x - 4x^3 = 0 \implies 2x(h^2 - 2x^2) = 0 \).
Since \( x > 0 \), we have:
\( h^2 - 2x^2 = 0 \implies x^2 = \frac{h^2}{2} \implies x = \frac{h}{\sqrt{2}} \).
Now, find \( y \):
\( y^2 = h^2 - x^2 = h^2 - \frac{h^2}{2} = \frac{h^2}{2} \implies y = \frac{h}{\sqrt{2}} \).
Since the second derivative \( \frac{d^2Z}{dx^2} = \frac{1}{4}(2h^2 - 12x^2) \) is negative at \( x^2 = h^2/2 \), this point yields the maximum area. Since \( x = y \), the triangle is isosceles.
Hence, proved.
In simple words: Maximize the square of the area formula to avoid square roots. Solving the derivative shows that the area is largest when the two legs of the triangle are equal.

Exam Tip: Maximizing the square of a function (like area) is a standard mathematical shortcut that avoids working with complex radical derivatives.

Question 2. A piece of wire 28(units) long is cut into two pieces. One piece is bent into the shape of a circle and other into the shape of a square. How should the wire be cut so that the combined area of the two figures is as small as possible.
Answer: Let the wire be cut into two pieces of lengths \( x \) and \( 28 - x \).
Let the piece of length \( x \) be bent into a circle of radius \( r \):
\( 2\pi r = x \implies r = \frac{x}{2\pi} \).
The area of this circle is:
\( A_1 = \pi r^2 = \pi \left( \frac{x}{2\pi} \right)^2 = \frac{x^2}{4\pi} \).

Let the second piece of length \( 28 - x \) be bent into a square of side \( s \):
\( 4s = 28 - x \implies s = \frac{28 - x}{4} \).
The area of this square is:
\( A_2 = s^2 = \left( \frac{28 - x}{4} \right)^2 = \frac{(28-x)^2}{16} \).

The combined area \( A \) is:
\( A = A_1 + A_2 = \frac{x^2}{4\pi} + \frac{(28-x)^2}{16} \).
Differentiating \( A \) with respect to \( x \):
\( \frac{dA}{dx} = \frac{2x}{4\pi} + \frac{2(28-x)(-1)}{16} = \frac{x}{2\pi} - \frac{28-x}{8} \).
Setting \( \frac{dA}{dx} = 0 \):
\( \frac{x}{2\pi} = \frac{28-x}{8} \implies 8x = 2\pi(28-x) \implies 4x = 28\pi - \pi x \)

\( \implies x(4 + \pi) = 28\pi \implies x = \frac{28\pi}{4 + \pi} \).
Since \( \frac{d^2A}{dx^2} = \frac{1}{2\pi} + \frac{1}{8} > 0 \), this point yields the minimum combined area.
Thus, the wire should be cut at a distance of \( \frac{28\pi}{4 + \pi} \) units from one end.
In simple words: Write the total area in terms of the cut position x, differentiate, set to zero, and show that the positive second derivative confirms this position minimizes the area.

Exam Tip: Always compute the second derivative to formally prove that your critical point represents a minimum (positive value) or a maximum (negative value).

Question 3. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Answer: Let \( 2x \) be the width and \( y \) be the height of the rectangular part of the window. The radius of the surmounting semicircle is \( x \).
The perimeter \( P \) of the window consists of the bottom width, two vertical sides, and the semicircular arc:
\( P = 2x + 2y + \pi x = 10 \)

\( \implies 2y = 10 - 2x - \pi x \implies y = 5 - x - \frac{\pi}{2}x \).
The total area \( A \) of the window (which determines the light admitted) is:
\( A = \text{Area of rectangle} + \text{Area of semicircle} \)

\( \implies A = 2xy + \frac{1}{2}\pi x^2 \).
Substituting \( y \):
\( A = 2x\left( 5 - x - \frac{\pi}{2}x \right) + \frac{1}{2}\pi x^2 = 10x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2 = 10x - 2x^2 - \frac{\pi}{2}x^2 \).
Differentiating \( A \) with respect to \( x \):
\( \frac{dA}{dx} = 10 - 4x - \pi x \).
Setting \( \frac{dA}{dx} = 0 \):
\( 10 - x(4 + \pi) = 0 \implies x = \frac{10}{4 + \pi} \).
Since \( \frac{d^2A}{dx^2} = -4 - \pi < 0 \), this value of \( x \) maximizes the area.
The dimensions of the window are:
Width: \( 2x = \frac{20}{4 + \pi} \text{ m} \).
Height of the rectangular part:
\( y = 5 - \left(\frac{10}{4+\pi}\right) - \frac{\pi}{2}\left(\frac{10}{4+\pi}\right) = \frac{20 + 5\pi - 10 - 5\pi}{4+\pi} = \frac{10}{4+\pi} \text{ m} \).
Thus, the dimensions are width \( \frac{20}{4+\pi} \text{ m} \) and height \( \frac{10}{4+\pi} \text{ m} \).
In simple words: Write the total area in terms of the width variable x, differentiate, set to zero to find the optimal width, and calculate the corresponding height.

Exam Tip: Choosing the width of the window as \( 2x \) instead of \( x \) makes the radius of the semicircle \( x \), which simplifies the algebraic fractions significantly.

Level III

Question 1. Find the area of the greatest isosceles triangle that can be inscribed in a given ellipse having its vertex coincident with one extremity of major axis.
Answer: Let the equation of the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Let one vertex of the inscribed isosceles triangle be at \( A(a, 0) \).
Let the other two vertices be \( B(x, y) \) and \( C(x, -y) \) on the ellipse, where \( x = a\cos \theta \) and \( y = b\sin \theta \).
The base of the isosceles triangle is \( BC = 2y = 2b\sin \theta \).
The altitude of the triangle is the distance from \( A(a, 0) \) to the line \( BC \) (which lies at \( x = a\cos \theta \)):
\( \text{Height} = a - x = a - a\cos \theta = a(1 - \cos \theta) \).
The area \( A \) of this triangle is:
\( A = \frac{1}{2} \cdot \text{Base} \cdot \text{Height} = \frac{1}{2} (2b\sin \theta) \cdot a(1 - \cos \theta) = ab\sin \theta(1 - \cos \theta) \).
Differentiating \( A \) with respect to \( \theta \):
\( \frac{dA}{d\theta} = ab \left[ \cos \theta(1 - \cos \theta) + \sin \theta(\sin \theta) \right] = ab \left[ \cos \theta - \cos^2 \theta + \sin^2 \theta \right] \)

\( \implies \frac{dA}{d\theta} = ab \left[ \cos \theta - \cos^2 \theta + 1 - \cos^2 \theta \right] = ab \left[ 1 + \cos \theta - 2\cos^2 \theta \right] \).
Setting \( \frac{dA}{d\theta} = 0 \):
\( 2\cos^2 \theta - \cos \theta - 1 = 0 \implies (2\cos \theta + 1)(\cos \theta - 1) = 0 \).
Since \( \cos \theta = 1 \) yields a trivial triangle of area 0, we choose:
\( 2\cos \theta + 1 = 0 \implies \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \).
At this angle, \( \sin \theta = \frac{\sqrt{3}}{2} \). The maximum area is:
\( A = ab\left(\frac{\sqrt{3}}{2}\right)\left(1 - \left(-\frac{1}{2}\right)\right) = ab\left(\frac{\sqrt{3}}{2}\right)\left(\frac{3}{2}\right) = \frac{3\sqrt{3}}{4}ab \).
Thus, the area of the greatest inscribed isosceles triangle is \( \frac{3\sqrt{3}}{4}ab \) square units.
In simple words: Represent the triangle's coordinates using parametric angles on the ellipse. Differentiate the area formula with respect to the angle, set it to zero, and solve to find the maximum area.

Exam Tip: Using parametric coordinates \( (a\cos \theta, b\sin \theta) \) is the most elegant way to solve optimization problems involving ellipses.

Question 2. An open box with a square base is to be made out of a given quantity of card board of area \(c^2\) square units. Show that the maximum volume of the box is \( \frac{c^3}{6\sqrt{3}} \) cubic units.[CBSE 2012 Comptt.]
Answer: Let \( x \) be the side of the square base and \( y \) be the height of the open box.
The surface area of the open box with a square base is:
\( S = x^2 + 4xy = c^2 \implies y = \frac{c^2 - x^2}{4x} \).
The volume \( V \) of the box is:
\( V = x^2 y = x^2 \left( \frac{c^2 - x^2}{4x} \right) = \frac{1}{4} (c^2 x - x^3) \).
Differentiating \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{1}{4} (c^2 - 3x^2) \).
Setting \( \frac{dV}{dx} = 0 \):
\( c^2 - 3x^2 = 0 \implies x^2 = \frac{c^2}{3} \implies x = \frac{c}{\sqrt{3}} \).
Since \( \frac{d^2V}{dx^2} = -\frac{3}{2}x < 0 \), this value of \( x \) maximizes the volume.
Substituting \( x = \frac{c}{\sqrt{3}} \) back into the volume formula:
\( V_{\max} = \frac{1}{4} \left[ c^2 \left(\frac{c}{\sqrt{3}}\right) - \left(\frac{c}{\sqrt{3}}\right)^3 \right] = \frac{1}{4} \left[ \frac{c^3}{\sqrt{3}} - \frac{c^3}{3\sqrt{3}} \right] = \frac{1}{4} \left[ \frac{2c^3}{3\sqrt{3}} \right] = \frac{c^3}{6\sqrt{3}} \) cubic units.
Hence, proved.
In simple words: Write the height in terms of the base side using the surface area constraint. Substitute this into the volume formula, differentiate, and solve to find the maximum volume.

Exam Tip: Be sure to write the surface area of the *open* box as \( x^2 + 4xy \), not \( 2x^2 + 4xy \), since there is no top lid.

Page 38

Question 3. A window is in the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window. [CBSE 2011]
Answer: Let \( x \) be the width and \( y \) be the height of the rectangular part of the window. The side of the surmounting equilateral triangle is \( x \).
The perimeter \( P \) consists of the bottom width, two vertical sides, and the two upper sides of the triangle:
\( P = x + 2y + 2x = 3x + 2y = 12 \)

\( \implies 2y = 12 - 3x \implies y = 6 - \frac{3}{2}x \).
The total area \( A \) of the window is:
\( A = \text{Area of rectangle} + \text{Area of equilateral triangle} \)

\( \implies A = xy + \frac{\sqrt{3}}{4}x^2 \).
Substituting \( y \):
\( A = x\left( 6 - \frac{3}{2}x \right) + \frac{\sqrt{3}}{4}x^2 = 6x - \frac{3}{2}x^2 + \frac{\sqrt{3}}{4}x^2 \).
Differentiating with respect to \( x \):
\( \frac{dA}{dx} = 6 - 3x + \frac{\sqrt{3}}{2}x \).
Setting \( \frac{dA}{dx} = 0 \):
\( 6 - x\left( 3 - \frac{\sqrt{3}}{2} \right) = 0 \implies x\left( \frac{6-\sqrt{3}}{2} \right) = 6 \implies x = \frac{12}{6-\sqrt{3}} \text{ m} \).
We rationalize the denominator:
\( x = \frac{12(6+\sqrt{3})}{36-3} = \frac{12(6+\sqrt{3})}{33} = \frac{4(6+\sqrt{3})}{11} \text{ m} \).
Now, find \( y \):
\( y = 6 - \frac{3}{2}\left[ \frac{12}{6-\sqrt{3}} \right] = 6 - \frac{18}{6-\sqrt{3}} = \frac{36 - 6\sqrt{3} - 18}{6-\sqrt{3}} = \frac{18 - 6\sqrt{3}}{6-\sqrt{3}} = \frac{6(3-\sqrt{3})}{6-\sqrt{3}} \text{ m} \).
Thus, the dimensions of the rectangular part are width \( \frac{4(6+\sqrt{3})}{11} \text{ m} \) and height \( \frac{6(3-\sqrt{3})}{6-\sqrt{3}} \text{ m} \).
In simple words: Express the height of the window in terms of its width using the perimeter limit. Write the total area, differentiate, and solve for the optimal dimensions.

Exam Tip: Rationalizing the denominator makes the final dimensions look much cleaner and easier to read.

Questions for self evaluation

Question 1. Sand is pouring from a pipe at the rate of 12 \(cm^3\)/s. 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 the height is 4 cm?
Answer: Let \( V \), \( r \), and \( h \) be the volume, radius, and height of the sand cone.
Given:
Rate of change of volume, \( \frac{dV}{dt} = 12 \text{ cm}^3\text{/s} \).
Height relationship, \( h = \frac{1}{6}r \implies r = 6h \log \).
The volume \( V \) of a cone is:
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6h)^2 h = 12\pi h^3 \).
Differentiating both sides with respect to time \( t \) using the chain rule:
\( \frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt} \).
Substituting \( \frac{dV}{dt} = 12 \) and \( h = 4 \text{ cm} \):
\( 12 = 36\pi (4)^2 \frac{dh}{dt} \)

\( \implies 12 = 576\pi \frac{dh}{dt} \)

\( \implies \frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi} \text{ cm/s} \).
Thus, the height of the sand cone is increasing at a rate of \( \frac{1}{48\pi} \text{ cm/s} \).
In simple words: Express the volume of the sand cone in terms of its height using the given ratio. Differentiate and solve for the rate of height growth when the height is 4 cm.

Exam Tip: Be sure to replace the radius with \( 6h \) before taking the derivative so you only have to work with a single variable \( h \).

Question 2. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?
Answer: Let the two equal sides of the isosceles triangle be \( a \). The base \( b \) is constant.
Given:
Rate of change of equal sides, \( \frac{da}{dt} = -3 \text{ cm/s} \) (negative because they are decreasing).
By dropping a perpendicular from the vertex to the base, the height \( h \) of the triangle is:
\( h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \).
The area \( A \) of the triangle is:
\( A = \frac{1}{2} \cdot \text{Base} \cdot \text{Height} = \frac{1}{2} b \sqrt{a^2 - \frac{b^2}{4}} \).
Differentiating \( A \) with respect to time \( t \):
\( \frac{dA}{dt} = \frac{1}{2} b \cdot \frac{1}{2\sqrt{a^2 - \frac{b^2}{4}}} \cdot 2a \frac{da}{dt} = \frac{ab \frac{da}{dt}}{2\sqrt{a^2 - \frac{b^2}{4}}} \).
We evaluate this rate when the equal sides are equal to the base, i.e., \( a = b \):
\( \frac{dA}{dt} = \frac{b^2 (-3)}{2\sqrt{b^2 - \frac{b^2}{4}}} = \frac{-3b^2}{2\sqrt{\frac{3b^2}{4}}} = \frac{-3b^2}{2\left(\frac{\sqrt{3}b}{2}\right)} = -\sqrt{3}b \text{ cm}^2\text{/s} \).
Thus, the area is decreasing at a rate of \( \sqrt{3}b \text{ cm}^2\text{/s} \).
In simple words: Find the height of the triangle using the Pythagorean theorem, write the area formula, differentiate with respect to time, and substitute the final values to solve.

Exam Tip: Be mindful of which variables are constant (like base \( b \)) and which are changing (like side \( a \)) before differentiating.

Question 3. Find the intervals in which the following function is strictly increasing or decreasing: f(x) = – \(2x^3\)– \(9x^2\)– 12x + 1
Answer: We first find the derivative of the function:
\( f'(x) = -6x^2 - 18x - 12 \).
Factoring out -6:
\( f'(x) = -6(x^2 + 3x + 2) = -6(x + 1)(x + 2) \).
To find critical points, set \( f'(x) = 0 \):
\( -6(x+1)(x+2) = 0 \implies x = -1 \) or \( x = -2 \).
These points divide the real line into three intervals: \( (-\infty, -2) \), \( (-2, -1) \), and \( (-1, \infty) \).

We test the sign of \( f'(x) \) in each interval:
1) On \( (-\infty, -2) \): Choose \( x = -3 \), then \( f'(-3) = -6(-2)(-1) = -12 < 0 \). Function is strictly decreasing.
2) On \( (-2, -1) \): Choose \( x = -1.5 \), then \( f'(-1.5) = -6(0.5)(-0.5) = 1.5 > 0 \). Function is strictly increasing.
3) On \( (-1, \infty) \): Choose \( x = 0 \), then \( f'(0) = -6(1)(2) = -12 < 0 \). Function is strictly decreasing.

Thus, the function is strictly increasing in \( (-2, -1) \) and strictly decreasing in \( (-\infty, -2) \cup (-1, \infty) \).
In simple words: Find the derivative, factor it, and identify the boundary points. Test points in each interval to find where the curve rises and where it falls.

Exam Tip: Keep track of the negative sign of the factored constant -6, as it reverses all interval signs compared to standard positive quadratics.

Question 4. Find the intervals in which the following function is strictly increasing or decreasing: f(x) = sinx + cosx , 0 ≤ x ≤ 2π
Answer: We differentiate \( f(x) = \sin x + \cos x \):
\( f'(x) = \cos x - \sin x \).
To find critical points, set \( f'(x) = 0 \):
\( \cos x - \sin x = 0 \implies \tan x = 1 \).
In the interval \( [0, 2\pi] \), the solutions are:
\( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
These points divide the domain into three intervals:

1) On \( \left[0, \frac{\pi}{4}\right) \): For example \( x = 0 \), \( f'(0) = \cos 0 - \sin 0 = 1 > 0 \). Strictly increasing.
2) On \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \): For example \( x = \pi \), \( f'(\pi) = \cos \pi - \sin \pi = -1 < 0 \). Strictly decreasing.
3) On \( \left(\frac{5\pi}{4}, 2\pi\right] \): For example \( x = \frac{3\pi}{2} \), \( f'\left(\frac{3\pi}{2}\right) = 0 - (-1) = 1 > 0 \). Strictly increasing.

Thus, the function is strictly increasing in \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \) and strictly decreasing in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).
In simple words: Find where the slope is zero, which happens at 45 and 225 degrees. Test angles in each region to determine where the function rises and falls.

Exam Tip: Verify that the endpoints are correctly included using square brackets where the interval is closed.

Question 5. For the curve y = \(4x^3\) – \(2x^5\), find all the points at which the tangent passes through the origin.
Answer: Let the point of contact be \( (x_1, y_1) \). This point lies on the curve, so:
\( y_1 = 4x_1^3 - 2x_1^5 \).
Differentiating the curve equation:
\( \frac{dy}{dx} = 12x^2 - 10x^4 \).
The slope of the tangent at \( (x_1, y_1) \) is \( 12x_1^2 - 10x_1^4 \).
The equation of the tangent line is:
\( y - y_1 = (12x_1^2 - 10x_1^4)(x - x_1) \).
Since this tangent line passes through the origin \( (0,0) \), we substitute \( x = 0 \) and \( y = 0 \):
\( -y_1 = -x_1(12x_1^2 - 10x_1^4) \implies y_1 = 12x_1^3 - 10x_1^5 \).
Equating the 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 \( x_1 = 0 \) or \( x_1 = \pm 1 \).
Now, we find the corresponding \( y \)-coordinates:
1) If \( x_1 = 0 \): \( y_1 = 0 \). Point is \( (0, 0) \).
2) If \( x_1 = 1 \): \( y_1 = 4(1)^3 - 2(1)^5 = 2 \). Point is \( (1, 2) \).
3) If \( x_1 = -1 \): \( y_1 = 4(-1)^3 - 2(-1)^5 = -2 \). Point is \( (-1, -2) \).
Thus, the points are \( (0, 0) \), \( (1, 2) \), and \( (-1, -2) \).
In simple words: Write the tangent equation at a generic point, substitute the origin, solve for x, and find the corresponding coordinates.

Exam Tip: Be sure to check all three roots of the factored equation to ensure you do not miss any points.

Question 6. Find the equation of the tangent line to the curve y = \(x^2\) – 2x +7 which is (a) parallel to the line 2x – y + 9 = 0 (b) perpendicular to the line 5y – 15x = 13.
Answer: Let the point of contact be \( (x_1, y_1) \).
Differentiating the curve: \( \frac{dy}{dx} = 2x - 2 \).
The slope of the tangent at \( (x_1, y_1) \) is \( m = 2x_1 - 2 \).

Part (a): Parallel to \( 2x - y + 9 = 0 \)
The slope of the line is 2. Since the tangent is parallel to this line:
\( 2x_1 - 2 = 2 \implies 2x_1 = 4 \implies x_1 = 2 \).
Find \( y_1 \): \( y_1 = 2^2 - 2(2) + 7 = 5 \). Point of contact is \( (2, 5) \).
Equation of the tangent: \( y - 5 = 2(x - 2) \implies 2x - y + 1 = 0 \).

Part (b): Perpendicular to \( 5y - 15x = 13 \)
The given line is \( 5y = 15x + 13 \implies y = 3x + \frac{13}{5} \), so its slope is 3.
Since the tangent is perpendicular to this line, the product of their slopes is -1:
\( (2x_1 - 2)(3) = -1 \implies 6x_1 - 6 = -1 \implies 6x_1 = 5 \implies x_1 = \frac{5}{6} \).
Find \( y_1 \): \( y_1 = \left(\frac{5}{6}\right)^2 - 2\left(\frac{5}{6}\right) + 7 = \frac{25}{36} - \frac{5}{3} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36} \).
Point of contact is \( \left(\frac{5}{6}, \frac{217}{36}\right) \).
Equation of the tangent: \( y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right) \)

\( \implies 36y - 217 = -12\left(x - \frac{5}{6}\right) = -12x + 10 \)

\( \implies 12x + 36y - 227 = 0 \).
In simple words: Find the slope of the curve using derivatives. For part (a), set it equal to the parallel line's slope. For part (b), set it equal to the negative reciprocal of the perpendicular line's slope.

Exam Tip: Remember the perpendicular condition \( m_1 m_2 = -1 \) when solving part (b).

Question 7. Prove that the curves x = \(y^2\) and xy = k cut at right angles if \(8k^2\) = 1.
Answer: Let the curves intersect at \( (x_1, y_1) \). This point satisfies both curves:
\( x_1 = y_1^2 \) and \( x_1 y_1 = k \implies y_1^3 = k \implies y_1 = k^{1/3} \).
Thus, \( x_1 = (k^{1/3})^2 = k^{2/3} \).

Differentiating the first curve \( x = y^2 \) with respect to \( x \):
\( 1 = 2y \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{2y} \).
The slope of the first tangent at \( (x_1, y_1) \) is \( m_1 = \frac{1}{2y_1} \).

Differentiating the second curve \( xy = k \) with respect to \( x \):
\( y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \).
The slope of the second tangent at \( (x_1, y_1) \) is \( m_2 = -\frac{y_1}{x_1} \).

For the curves to cut at right angles (orthogonally):
\( m_1 m_2 = -1 \)

\( \implies \left(\frac{1}{2y_1}\right) \left(-\frac{y_1}{x_1}\right) = -1 \)

\( \implies -\frac{1}{2x_1} = -1 \implies 2x_1 = 1 \implies x_1 = \frac{1}{2} \).
Substituting \( x_1 = k^{2/3} \):
\( 2k^{2/3} = 1 \).
Cubing both sides of the equation:
\( (2k^{2/3})^3 = 1^3 \implies 8k^2 = 1 \).
Hence, proved.
In simple words: Find the point of intersection. Calculate both slopes using derivatives, multiply them to get -1, and show that this simplifies to the given condition \( 8k^2 = 1 \).

Exam Tip: Cubing both sides of \( 2k^{2/3} = 1 \) is the key algebraic step to clear the fractional exponent and complete the proof.

Question 8. Using differentials, find the approximate value of each of the following up to 3places of decimal : (i) \( (26)^{1/3} \) (ii) \( (32.15)^{1/5} \)
Answer: We approximate both values:
Part (i): \( (26)^{1/3} \)
Let \( f(x) = x^{1/3} \). Choose \( x = 27 \) (perfect cube) and \( \Delta x = -1 \).
\( f'(x) = \frac{1}{3x^{2/3}} \).
\( f(x+\Delta x) \approx f(x) + f'(x)\Delta x \)

\( \implies (26)^{1/3} \approx (27)^{1/3} + \frac{1}{3(27)^{2/3}}(-1) \)

\( \implies (26)^{1/3} \approx 3 - \frac{1}{27} \approx 3 - 0.037 = 2.963 \).

Part (ii): \( (32.15)^{1/5} \)
Let \( f(x) = x^{1/5} \). Choose \( x = 32 \) (since \( 2^5 = 32 \)) and \( \Delta x = 0.15 \).
\( f'(x) = \frac{1}{5x^{4/5}} \).
\( f(x+\Delta x) \approx f(x) + f'(x)\Delta x \)

\( \implies (32.15)^{1/5} \approx (32)^{1/5} + \frac{1}{5(32)^{4/5}}(0.15) \)

\( \implies (32.15)^{1/5} \approx 2 + \frac{0.15}{5(16)} = 2 + \frac{0.15}{80} = 2 + 0.001875 \approx 2.002 \).
Thus, the approximate values are (i) 2.963 and (ii) 2.002.
In simple words: Find nearby perfect powers (27 for cube root, 32 for fifth root), calculate their derivatives, and use differentials to estimate the small changes.

Exam Tip: Be precise with decimal places; rounding to three decimal places is a standard requirement for these questions.

Question 9. 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 constant radius of the sphere. Let \( r \) be the radius and \( h \) be the height of the inscribed cone.
Let \( x \) be the distance from the center of the sphere to the base of the cone. Thus, the height of the cone is:
\( h = R + x \).
From the right triangle inside the sphere:
\( r^2 + x^2 = R^2 \implies r^2 = R^2 - x^2 \).
The volume \( V \) of the cone is:
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (R^2 - x^2)(R + x) = \frac{1}{3}\pi (R^3 + R^2 x - Rx^2 - x^3) \).
Differentiating \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{1}{3}\pi (R^2 - 2Rx - 3x^2) \).
Setting \( \frac{dV}{dx} = 0 \):
\( 3x^2 + 2Rx - R^2 = 0 \implies (3x - R)(x + R) = 0 \).
Since \( x > -R \), we choose:
\( x = \frac{R}{3} \).
The second derivative \( \frac{d^2V}{dx^2} = \frac{1}{3}\pi(-2R-6x) < 0 \), verifying that \( x = R/3 \) maximizes the volume.
The height is \( h = R + \frac{R}{3} = \frac{4R}{3} \), and \( r^2 = R^2 - \frac{R^2}{9} = \frac{8R^2}{9} \).
The maximum volume of the cone is:
\( V_{\text{cone}} = \frac{1}{3}\pi \left( \frac{8R^2}{9} \right) \left( \frac{4R}{3} \right) = \frac{32}{81}\pi R^3 = \frac{8}{27} \left( \frac{4}{3}\pi R^3 \right) = \frac{8}{27} V_{\text{sphere}} \).
Hence, proved.
In simple words: Express the cone's dimensions using the sphere's radius. Substitute into the cone's volume formula, differentiate, and show that the maximum volume is exactly 8/27 of the sphere's volume.

Exam Tip: Defining \( x \) as the distance from the center of the sphere to the cone's base is a very neat trick that makes the algebra simple.

Question 10. An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.
Answer: Let \( x \) be the side of the square cut from each corner. The dimensions of the open box will be:
Length: \( 8 - 2x \)
Width: \( 3 - 2x \)
Height: \( x \)
Since the width must be positive, \( 3 - 2x > 0 \implies x < 1.5 \).
The volume \( V \) of the box is:
\( V = x(8 - 2x)(3 - 2x) = x(24 - 22x + 4x^2) = 4x^3 - 22x^2 + 24x \).
Differentiating with respect to \( x \):
\( \frac{dV}{dx} = 12x^2 - 44x + 24 \).
Setting \( \frac{dV}{dx} = 0 \):
\( 4(3x^2 - 11x + 6) = 0 \implies 4(3x - 2)(x - 3) = 0 \).
This gives \( x = \frac{2}{3} \) or \( x = 3 \).
Since \( x < 1.5 \), we choose \( x = \frac{2}{3} \text{ m} \).
Checking the second derivative:
\( \frac{d^2V}{dx^2} = 24x - 44 \). At \( x = 2/3 \), \( \frac{d^2V}{dx^2} = 16 - 44 = -28 < 0 \), confirming a maximum.
The maximum volume is:
\( V = \left(\frac{2}{3}\right) \left( 8 - \frac{4}{3} \right) \left( 3 - \frac{4}{3} \right) = \left(\frac{2}{3}\right) \left(\frac{20}{3}\right) \left(\frac{5}{3}\right) = \frac{200}{27} \text{ m}^3 \approx 7.41 \text{ m}^3 \).
Thus, the volume of the largest box is \( \frac{200}{27} \text{ m}^3 \).
In simple words: Write the volume of the folded box in terms of the corner cut length x, differentiate, find the valid root, and calculate the maximum volume.

Exam Tip: Always make sure your optimal value \( x \) satisfies the physical boundary constraints of the sheet's width.

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