JEE Main 202125 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 25 July · Morning Shift · Mathematics · Application of Derivatives

Let f(x) = 3 ^ 4 x + 10 ^ 3 x + 6 ^ 2 x - 3 , x ∈ [- (π)/(6),...

Let

f\left(x\right) = 3\sin^{4}x + 10\sin^{3}x + 6\sin^{2}x - 3

x \in \left[- \dfrac{\pi}{6}, \dfrac{\pi}{2}\right]

. Then,

f

is :

Let f(x) = 3 ^ 4 x + 10 ^ 3 x + 6 ^ 2 x - 3 , x ∈ [- (π)/(6), (π)/(2)] . Then, f is :

Options

Aincreasing in $\left(- \dfrac{\pi}{6}, \dfrac{\pi}{2}\right)$
Bdecreasing in $\left(0, \dfrac{\pi}{2}\right)$
Cincreasing in $\left(- \dfrac{\pi}{6}, 0\right)$
Ddecreasing in $\left(- \dfrac{\pi}{6}, 0\right)$

Correct answer

D. decreasing in $\left(- \dfrac{\pi}{6}, 0\right)$

Step-by-step solution

$f\left(x\right) = 3\sin^{4}x + 10\sin^{3}x + 6\sin^{2}x - 3, x \in \left[- \dfrac{\pi}{6}, \dfrac{\pi}{2}\right]$

$⇒ {f}^{'}\left(x\right) = 12\sin^{3}xcosx + 30\sin^{2}xcosx + 12\sin x \cos x$

$= 6\sin x \cos x\left(2\sin^{2}x + 5\sin x + 2\right)$

$= 6\sin x \cos x ( 2\sin x + 1 ) ( \sin x + 2 )$

Let f(x) = 3 ^ 4 x + 10 ^ 3 x + 6 ^ 2 x - 3 , x ∈ [- (π)/(6), (π)/(2)] . Then, f is :

From wavy curve, we can say $f ' ( x ) < 0$ in \left(- \dfrac{\pi}{6}, 0\right) $</p><p>So, the function is Decreasing in \left(- \dfrac{\pi}{6}, 0\right)$

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