The interval in which the function f(x)=x^3-12x^2+36x+20 is increasing, is Option 1: (-∞,-2)(6,∞) Option 2: [-∞,-3)(2,∞) Option 3: (-∞,1](3,∞) Option 4: (-∞,2)(6,∞)

Answer: (-∞,2)(6,∞). Solution: f'(x)=3x^2-24x+36=3(x-2)(x-6), which is positive when x 6. So f increases on (-∞,2)(6,∞).

Calculus

Application of Derivatives

Easy

applied

The interval in which the function $f(x)=x^3-12x^2+36x+20$ is increasing, is

$(-\infty,-2)\cup(6,\infty)$

$[-\infty,-3)\cup(2,\infty)$

$(-\infty,1]\cup(3,\infty)$

$(-\infty,2)\cup(6,\infty)$

✅ Correct Option: 4

$f'(x)=3x^2-24x+36=3(x-2)(x-6)$ , which is positive when $x<2$ or $x>6$ . So $f$ increases on $(-\infty,2)\cup(6,\infty)$ .

2025: 21 May Shift 1

Match List-I with List-II

List-I	List-II
(A) Maximum value of $f(x) = \sin^2 x - \cos^2 x$ , $\forall x \in (\pi, 2\pi)$ is	(I) 0
(B) Minimum value of $f(x) = \sin x \cos x$	(II) 1
(C) Point of Minima of $f(x) = x^x$ $(x > 0)$	(III) $-\frac{1}{2}$
(D) Maximum value of $f(x) = -x^{2026}$	(IV) $\frac{1}{e}$

Choose the correct answer from the options given below:

2025: 29 May Shift 2

The rate of change of volume of a sphere with respect to its surface area, when the radius is 6cm is:

2025: 14 May Shift 1

Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) The minimum value of } f(x) = (2x - 1)^2 + 3 & \text{(I) } 4 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) The maximum value of } f(x) = -|x + 1| + 4 & \text{(II) } 10 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) The minimum value of } f(x) = \sin(2x) + 6 & \text{(III) } 3 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) The maximum value of } f(x) = -(x - 1)^2 + 10 & \text{(IV) } 5 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

2022: 16 July Shift 2

The interval in which the function given by $f(x) = x^2 e^{-x}$ is strictly increasing is:

2023: 30 May Shift 3

The slope of the tangent to the curve $x = at^2$ , $y = 2at$ at 't' is :

2026: 19 May Shift 2

The number of points of the local maximum value of the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 2$ is,

The interval in which the function f(x)=x3−12x2+36x+20f(x)=x^3-12x^2+36x+20f(x)=x3−12x2+36x+20 is increasing, is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The interval in which the function $f(x)=x^3-12x^2+36x+20$ is increasing, is