Calculus

Application of Derivatives

Easy

common

For the function $f(x) = 2x^3 - 9x^2 + 12x - 5$ , $x \in [0, 3]$ , match List-I with List-II :

List-I List-II
A. Absolute maximum value (I) $3$
B. Absolute minimum value (II) $0$
C. Point of maxima (III) $-5$
D. Point of minima (IV) $4$

List-I	List-II
A. Absolute maximum value	(I) $3$
B. Absolute minimum value	(II) $0$
C. Point of maxima	(III) $-5$
D. Point of minima	(IV) $4$

(A) - (IV), (B) - (II), (C) - (I), (D) - (III)

(A) - (II), (B) - (III), (C) - (I), (D) - (IV)

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

(A) - (IV), (B) - (III), (C) - (I), (D) - (II)

✅ Correct Option: 4

To find absolute max/min on a closed interval, we need to evaluate $f(x)$ at all critical points and endpoints.

Find critical points by setting $f'(x) = 0$ :

$f'(x) = 6x^2 - 18x + 12 = 0$

$x^2 - 3x + 2 = 0$

$(x - 1)(x - 2) = 0$

$x = 1$ and $x = 2$

Both lie within $[0, 3]$ , so both are valid.

Evaluate $f(x)$ at $x = 0, 1, 2, 3$ :

$f(0) = 2(0)^3 - 9(0)^2 + 12(0) - 5 = -5$

$f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0$

$f(2) = 2(8) - 9(4) + 12(2) - 5 = 16 - 36 + 24 - 5 = -1$

$f(3) = 2(27) - 9(9) + 12(3) - 5 = 54 - 81 + 36 - 5 = 4$

From these values:

Absolute maximum value $= 4$ , occurring at $x = 3$ (point of maxima)
Absolute minimum value $= -5$ , occurring at $x = 0$ (point of minima)

Note: Even though $x = 1$ and $x = 2$ are critical points, the function reaches its highest and lowest values at the endpoints of the interval. This is why we always check endpoints on a closed interval.

A $\to$ (IV), B $\to$ (III), C $\to$ (I), D $\to$ (II)

2026: 25 May Shift 1

The points in the interval $[0,2\pi]$ , where the function $f(x) = \cos2x + 2\sin x\cos x$ , attains its local minima, is/are:

A. $\pi/8$

B. $5\pi/8$

C. $13\pi/8$

D. $11\pi/8$

Choose the correct answer from the options given below:

2025: 21 May Shift 2

Match List-I with List-II

List-I (Curve)	List-II (Slope of tangent at $x = 4$ )
(A) $y = \sqrt{x^3}$	(I) -1
(B) $y = \sqrt{x}$	(II) 1
(C) $y = x^3 - 47x$	(III) 1/4
(D) $xy = 16$	(IV) 3

Choose the correct answer from the options given below:

2026: 31 May Shift 2

The rate of change of $\sqrt{x^2-2}$ with respect to $\dfrac{x}{x+1}$ at $x=2$ is :

2023: 30 May Shift 3

The interval in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly decreasing is :

2026: 26 May Shift 2

Consider the function $f(x) = x^4 - 2x^2$ , then which of the following are TRUE?

A. $f(x)$ is increasing on $(-1,0)$

B. $f(x)$ is decreasing on $(0,1)$

C. $f(x)$ is increasing on $(-\infty,-1)$

D. $f(x)$ is decreasing on $(1,\infty)$

Choose the correct answer from the options given below:

2025: 22 May Shift 2

Which of the following statements are true?

(A) The function $f(x) = \frac{x^4}{4} - \frac{4}{3}x^3 + \frac{x^2}{2} + 6x$ has 3 critical points.

(B) The function $f(x) = |x| + 3$ has no minimum value.

(D) $f(x) = x^2$ has minima at $x=0$ .

Choose the correct answer from the options given below:

For the function f(x)=2x3−9x2+12x−5f(x) = 2x^3 - 9x^2 + 12x - 5f(x)=2x3−9x2+12x−5, x∈[0,3]x \in [0, 3]x∈[0,3], match List-I with List-II : List-IList-IIA. Absolute maximum value(I) 333B. Absolute minimum value(II) 000C. Point of maxima(III) −5-5−5D. Point of minima(IV) 444

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

For the function $f(x) = 2x^3 - 9x^2 + 12x - 5$ , $x \in [0, 3]$ , match List-I with List-II :

List-I List-II
A. Absolute maximum value (I) $3$
B. Absolute minimum value (II) $0$
C. Point of maxima (III) $-5$
D. Point of minima (IV) $4$