CUET UG 2025 — Mathematics Calculus

Given: $f(x) = x^3 - 3x$f(x)=x3−3x

The first derivative is:

$$f'(x) = 3x^2 - 3$$f′(x)=3x2−3

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

$$3x^2 - 3 = 0$$3x2−3=0

$$3x^2 = 3$$3x2=3

$$x^2 = 1$$x2=1

$x = 1$x=1 or $x = -1$x=−1

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The second derivative is:

$$f''(x) = 6x$$f′′(x)=6x

At $x = -1$x=−1:

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

This is a local maximum point.

At $x = 1$x=1:

$$f''(1) = 6(1) = 6 > 0$$f′′(1)=6(1)=6>0

This is a local minimum point.

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Local maximum value at $x = -1$x=−1:

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

$$f(-1) = -1 + 3$$f(−1)=−1+3

$$f(-1) = 2$$f(−1)=2

Local minimum value at $x = 1$x=1:

$$f(1) = (1)^3 - 3(1)$$f(1)=(1)3−3(1)

$$f(1) = 1 - 3$$f(1)=1−3

$$f(1) = -2$$f(1)=−2

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Matching:

List-I	Answer	List-II
(A) Point of local Maxima	$x = -1$x=−1	(II)
(B) Point of local Minima	$x = 1$x=1	(I)
(C) Local maximum value	$2$2	(III)
(D) Local minimum value	$-2$−2	(IV)

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