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CUET Mathematics 2025 PYQs13 May Shift 1. Free, no login required.

Calculus

past year questions, CUET Mathematics 2025 13 May Shift 1

Application of Derivatives

past year questions, Calculus, CUET Mathematics 2025 13 May Shift 1

Medium

core

Consider the function f(x)=x3−3xf(x) = x^3 - 3x. Then

Match List-I with List-II

List-IList-II
(A) Point of local Maxima(I) 1
(B) Point of local Minima(II) -1
(C) Local maximum value(III) 2
(D) Local minimum value(IV) -2

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

Given: f(x)=x3−3xf(x) = x^3 - 3x

The first derivative is:

f′(x)=3x2−3f'(x) = 3x^2 - 3

To find critical points, set f′(x)=0f'(x) = 0:

3x2−3=03x^2 - 3 = 0

3x2=33x^2 = 3

x2=1x^2 = 1

x=1x = 1 or x=−1x = -1

The second derivative is:

f′′(x)=6xf''(x) = 6x

At x=−1x = -1:

f′′(−1)=6(−1)=−6<0f''(-1) = 6(-1) = -6 < 0

This is a local maximum point.

At x=1x = 1:

f′′(1)=6(1)=6>0f''(1) = 6(1) = 6 > 0

This is a local minimum point.

Local maximum value at x=−1x = -1:

f(−1)=(−1)3−3(−1)f(-1) = (-1)^3 - 3(-1)

f(−1)=−1+3f(-1) = -1 + 3

f(−1)=2f(-1) = 2

Local minimum value at x=1x = 1:

f(1)=(1)3−3(1)f(1) = (1)^3 - 3(1)

f(1)=1−3f(1) = 1 - 3

f(1)=−2f(1) = -2

Matching:

List-IAnswerList-II
(A) Point of local Maximax=−1x = -1(II)
(B) Point of local Minimax=1x = 1(I)
(C) Local maximum value22(III)
(D) Local minimum value−2-2(IV)

Therefore, the correct answer is: (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

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