CBSE Class 12 Mathematics Application Of Derivatives MCQs Set 08

Assertion and Reason Questions

Directions: In the context of above two statements, which one of the following is correct?
(a) Both (A) and (B) are true and R is the correct explanation A.
(b) Both (A) and (R) are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A) : The points of contact of the vertical tangents to \( x = 2 - 3 \sin \theta \), \( y = 3 + 2 \cos \theta \) are (– 1, 3) and (5, 3).
Reason (R) : For vertical tangent, \( \frac{dx}{d\theta} = 0 \).
Answer: (a) Both (A) and (B) are true and R is the correct explanation A.

Question. Assertion (A) : If \( y^2 = 3 + 2x - x^2 \) then, at (3, 0) and (– 1, 0) tangent is perpendicular to x-axis.
Reason (R) : At (3, 0) and (– 1, 0), \( \frac{dy}{dx} = \infty \).
Answer: (a) Both (A) and (B) are true and R is the correct explanation A.

Question. Assertion (A) : The points on the curve \( y^2 = x + \sin x \) at which the tangent is parallel to x-axis lie on a striaght line.
Reason (R) : Tangent is parallel to x-axis, then \( \frac{dy}{dx} = 0 \) or \( \frac{dy}{dx} = \infty \).
Answer: (c) A is true but R is false.

Question. Assertion (A) : Equation of tangents to the curve \( f(x) = x^2 \) at the point where slope of tangent is equal to functional value of the curve is \( 4x - y - 4 = 0, y = 0 \).
Reason (R) : \( f'(x) = f(x) \), at functional value of curve.
Answer: (a) Both (A) and (B) are true and R is the correct explanation A.

Case Study 1

A new room is to be constructed in a house to increase the living area into it. A window is to be opened on one of the walls of the room in shape of rectangle surmounted by an equilateral triangle. If the perimeter of window is 12 m and they want to bring maximum light from the window.

Question. Express y in terms of x :
(a) \( y = 6 - \frac{3}{2} x \)
(b) \( y = 3x - 6 \)
(c) \( y = \frac{3}{2} x - 12 \)
(d) \( y = \frac{3}{2} x - 6 \)
Answer: (a) \( y = 6 - \frac{3}{2} x \)

Question. If A represent the area of window, what is the function of in terms of x ?
(a) \( A = \frac{x - 3x^2 + \sqrt{3}x^2}{4} \)
(b) \( A = \frac{3x^2}{2} - \frac{\sqrt{3}}{4} x^2 - 6x \)
(c) \( A = 6x - \frac{3}{2} x^2 + \frac{\sqrt{3}}{4} x^2 \)
(d) \( A = (\frac{3+\sqrt{3}}{4}) x^2 + 6x \)
Answer: (c) \( A = 6x - \frac{3}{2} x^2 + \frac{\sqrt{3}}{4} x^2 \)

Question. For what value of x, the area will be maximum ?
(a) \( \frac{12}{6 + \sqrt{3}} \)
(b) \( \frac{12}{6 - \sqrt{3}} \)
(c) \( \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \)
(d) \( \frac{6-\sqrt{3}}{12} \)
Answer: (b) \( \frac{12}{6 - \sqrt{3}} \)

Question. Write the breadth of window :
(a) \( \frac{6 + \sqrt{3}}{12} \)
(b) \( \frac{6 - \sqrt{3}}{12} \)
(c) \( \frac{12}{6 + \sqrt{3}} \)
(d) \( \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \)
Answer: (d) \( \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \)

Question. What is the maximum area of triangular part of window ?
(a) \( \frac{63\sqrt{3}}{39 - 17\sqrt{3}} \)
(b) \( \frac{36\sqrt{3}}{39 - 12\sqrt{3}} \)
(c) \( \frac{39 - 12\sqrt{3}}{36\sqrt{3}} \)
(d) \( \frac{36\sqrt{3}}{39 + 12\sqrt{3}} \)
Answer: (b) \( \frac{36\sqrt{3}}{39 - 12\sqrt{3}} \)

Case Study 2

Vidhan Sabha is situated in Delhi along the curve \( y = x^2 + 7 \). A security guard is standing at a point (1, 8) on the curve who keeps his eyes horizontally and vertically on the people for security purposes. Security guard has a doubt on a person who is standing at a point (3, 7). He catches the person after a while.

Question. Find the slope of tangent at given points :
(a) 2
(b) \( -\frac{1}{2} \)
(c) \( \frac{1}{2} \)
(d) – 2
Answer: (a) 2

Question. Find the slope of normal to the curve at given point :
(a) 2
(b) \( -\frac{1}{2} \)
(c) \( \frac{1}{2} \)
(d) – 2
Answer: (b) \( -\frac{1}{2} \)

Question. Find the equation of tangent to the given curve :
(a) \( y = 2x - 6 \)
(b) \( y = x + 6 \)
(c) \( 2y = x - 6 \)
(d) \( y = 2x + 6 \)
Answer: (d) \( y = 2x + 6 \)

Question. Find the equation of normal to the curve :
(a) \( x + 2y = 17 \)
(b) \( x - 2y = 17 \)
(c) \( x - 2y + 17 = 0 \)
(d) \( x + 2y + 7 = 0 \)
Answer: (a) \( x + 2y = 17 \)

Question. How much distance covered by security guard to catch that person ?
(a) \( \frac{1}{\sqrt{5}} \)
(b) 5
(c) \( \sqrt{5} \)
(d) 4
Answer: (c) \( \sqrt{5} \)

Case Study 3

After the lockdown period of Covid-19, it is difficult to restart the business of Touring Agency. A tour operator charges Rs. 200 per passenger for 50 passengers. He is giving a discount of Rs. 5 for each 10 passenger in excess of 50. He wants to maximise his earning during this pandemic period.

Question. Let x be the number of passengers and y is the earning per passenger of the operator. Write the function of y in terms of x.
(a) \( y = 225 - \frac{x}{2} \)
(b) \( y = \frac{x}{2} - 225 \)
(c) \( y = 200 + \frac{x}{2} \)
(d) \( y = 200 - \frac{x}{2} \)
Answer: (a) \( y = 225 - \frac{x}{2} \)

Question. Give the slope of that earning function.
(a) \( -\frac{1}{5} \)
(b) \( \frac{1}{10} \)
(c) \( \frac{1}{2} \)
(d) \( -\frac{1}{2} \)
Answer: (d) \( -\frac{1}{2} \)

Question. What is the total earning function ?
(a) \( 225 - \frac{x}{2} \)
(b) \( 225x - \frac{x^2}{2} \)
(c) \( 220 - \frac{x}{2} \)
(d) \( 220 - \frac{x^2}{3} \)
Answer: (b) \( 225x - \frac{x^2}{2} \)

Question. At what value of x he can have the maximum amount ?
(a) 200
(b) 225
(c) 175
(d) 220
Answer: (b) 225

Question. What is his maximum amount of earning ?
(a) 225
(b) 175
(c) 25000
(d) 25312.50
Answer: (d) 25312.50