Function f(x)=3x, x [0,()/2] is decreasing in the interval Option 1: [0,()/2] Option 2: [0,()/6] Option 3: [()/6,()/2] Option 4: [0,()/4]

Answer: [()/6,()/2]. Solution: f'(x)=33x, which is negative when 2<3x<(3)/2, i.e. 6<x<2. So f decreases on [6,2].

Calculus

Application of Derivatives

Medium

core

Function $f(x)=\sin3x, x \in \left[0,\frac{\pi}{2}\right]$ is decreasing in the interval

$\left[0,\frac{\pi}{2}\right]$

$\left[0,\frac{\pi}{6}\right]$

$\left[\frac{\pi}{6},\frac{\pi}{2}\right]$

$\left[0,\frac{\pi}{4}\right]$

✅ Correct Option: 3

$f'(x)=3\cos3x$ , which is negative when $\frac\pi2<3x<\frac{3\pi}{2}$ , i.e. $\frac\pi6<x<\frac\pi2$ . So $f$ decreases on $\left[\frac\pi6,\frac\pi2\right]$ .

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Function f(x)=sin⁡3x,x∈[0,π2]f(x)=\sin3x, x \in \left[0,\frac{\pi}{2}\right]f(x)=sin3x,x∈[0,2π​] is decreasing in the interval

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CUET Mathematics Past Year Questions (Topic-Wise)

Function $f(x)=\sin3x, x \in \left[0,\frac{\pi}{2}\right]$ is decreasing in the interval