Calculus

Application of Derivatives

Easy

common

Function $f(x) = x + \frac{1}{x}, x \neq 0$ is increasing in the interval

$(-1, 0)$

$(0, 1)$

$(-1, 1)$

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

✅ Correct Option: 4

Here $f'(x)=1-\frac{1}{x^2}$ , which is positive exactly when $x^2>1$ , that is $|x|>1$ . Hence the function increases on $(-\infty,-1)\cup(1,\infty)$ .

2025: 14 May Shift 2

Match List-I with List-II

The function $f(x) = 2x^3 - 15x^2 + 36x + 5$ for $x \in [2,5]$ has

List-I	List-II
(A) absolute maximum value	(I) 5
(B) absolute minimum value	(II) 60
(C) point of absolute maxima	(III) 3
(D) point of absolute minima	(IV) 32

Choose the correct answer from the options given below:

2025: 3 June Shift 1

The function $f(x) = 4 - 3x + 3x^2 - x^3$ is (Here $\mathbb{R}$ is set of real numbers)

2026: 18 May Shift 1

If $y = a\log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$ , then

2025: 3 June Shift 1

Let f be a function defined by $f(x) = 2x^3 - 3x^2 - 36x + 2$ , then which of the following are correct?

(A) The critical points of f(x) are -2 and 3.

(B) The function f(x) increases in the interval $(3, \infty)$

(D) The function f(x) increases in the interval (-2,3)

Choose the correct answer from the options given below:

2023: 15 June Shift 2

The interval in which the function $f(x) = 10 - 6x - 2x^2$ is decreasing is :

2023: 22 May Shift 3

Interval in which the function $f(x) = 2x^3 - 3x^2 - 12x + 10$ is decreasing is :

Function f(x)=x+1x,x≠0f(x) = x + \frac{1}{x}, x \neq 0f(x)=x+x1​,x=0 is increasing in the interval

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Function $f(x) = x + \frac{1}{x}, x \neq 0$ is increasing in the interval