Interval in which the function f given by f(x) = x - 4x, x (0, ()/2) is strictly decreasing is Option 1: 0 x > 0

Answer: 0 1/2 The value x = 1/2 at x = ()/3 Since cosine decreases from 1 to 0 as x goes from 0 to ()/2, the inequality x > 1/2 holds when x < ()/3. Combined with the given domain: 0 < x < ()/3 Therefore, the function is strictly decreasing in the interval (0, ()/3).

Calculus

Application of Derivatives

Medium

core

Interval in which the function $f$ given by $f(x) = \tan x - 4x$ , $x \in (0, \frac{\pi}{2})$ is strictly decreasing is

$0 < x < \frac{\pi}{3}$

$\frac{\pi}{3} < x < \frac{\pi}{2}$

$\frac{\pi}{4} < x < \frac{\pi}{2}$

$\frac{\pi}{2} > x > 0$

✅ Correct Option: 1

Given: $f(x) = \tan x - 4x$ where $x \in (0, \frac{\pi}{2})$

A function is strictly decreasing when its derivative is negative.

Taking the derivative:

$f'(x) = \sec^2 x - 4$

For the function to be strictly decreasing:

$f'(x) < 0$

$\sec^2 x - 4 < 0$

$\sec^2 x < 4$

Since $\sec^2 x = \frac{1}{\cos^2 x}$ :

$\frac{1}{\cos^2 x} < 4$

$1 < 4\cos^2 x$

$\cos^2 x > \frac{1}{4}$

Taking the square root (note that $\cos x$ is positive in $(0, \frac{\pi}{2})$ ):

$\cos x > \frac{1}{2}$

The value $\cos x = \frac{1}{2}$ at $x = \frac{\pi}{3}$

Since cosine decreases from 1 to 0 as $x$ goes from 0 to $\frac{\pi}{2}$ , the inequality $\cos x > \frac{1}{2}$ holds when $x < \frac{\pi}{3}$ .

Combined with the given domain:

$0 < x < \frac{\pi}{3}$

Therefore, the function is strictly decreasing in the interval $(0, \frac{\pi}{3})$ .

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Function $f(x) = x^3 - 3x + 3$ is

(A) Increasing in the interval $(-1, 1)$

(B) Increasing in the interval $(1, \infty)$

(D) Increasing in the interval $(-\infty, -1) \cup (1, \infty)$

Choose the correct answer from the options given below:

Interval in which the function fff given by f(x)=tan⁡x−4xf(x) = \tan x - 4xf(x)=tanx−4x, x∈(0,π2)x \in (0, \frac{\pi}{2})x∈(0,2π​) is strictly decreasing is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Interval in which the function $f$ given by $f(x) = \tan x - 4x$ , $x \in (0, \frac{\pi}{2})$ is strictly decreasing is