The interval in which f(x) = x/2 + 2/x is a decreasing function of x is : Option 1: [-2, 2] Option 2: (-∞, -2) (2, ∞) Option 3: (-2, 0) (0, 2) Option 4: (-2, 2)

Answer: (-2, 0) (0, 2). Solution: f'(x) = 1/2 - 2/x^2 = (x^2 - 4)/2x^2. For decreasing, f'(x) < 0 x^2 - 4 < 0 -2 < x < 2, x ≠ 0. So decreasing on (-2, 0) (0, 2).

Calculus

Application of Derivatives

Medium

The interval in which $f(x) = \frac{x}{2} + \frac{2}{x}$ is a decreasing function of $x$ is :

$[-2, 2]$

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

$(-2, 0) \cup (0, 2)$

$(-2, 2)$

✅ Correct Option: 3

$f'(x) = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2}$ . For decreasing, $f'(x) < 0 \Rightarrow x^2 - 4 < 0 \Rightarrow -2 < x < 2$ , $x \neq 0$ . So decreasing on $(-2, 0) \cup (0, 2)$ .

2026: 12 May Shift 2

If $x$ is real, then the minimum value of $x^2 - 8x + 25$ is

2025: 15 May Shift 1

If the function $f(x) = 2x^3 + 9x^2 + 12x-1$ is given,then $f(x)$ have

2026: 22 May Shift 2

The interval in which $f(x)=\frac{x}{\log x}$ is increasing, is:

2025: 22 May Shift 1

Match List-I with List-II

List-I	List-II
Function	Increasing on the interval
(A) $f(x) = -x^2 - 2x + 1$	(I) $(-\infty, -1)$
(B) $f(x) = x^2 + 1$	(II) $(1, \infty)$
(C) $f(x) = x^2 - 2x + 3$	(III) $(-\infty, 0)$
(D) $f(x) = -x^2$	(IV) $(0, \infty)$

Choose the correct answer from the options given below:

2026: 29 May Shift 1

The local maximum value of the function $f(x)=\dfrac{b^2}{a^4}(a^2x^2-x^4), a,b>0$ is equal to:

2025: 13 May Shift 2

The absolute maximum value of the function $f(x) = 4x - \frac{1}{2}x^2$ in the interval $\left[-2, \frac{9}{2}\right]$ is

The interval in which f(x)=x2+2xf(x) = \frac{x}{2} + \frac{2}{x}f(x)=2x​+x2​ is a decreasing function of xxx is :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The interval in which $f(x) = \frac{x}{2} + \frac{2}{x}$ is a decreasing function of $x$ is :