JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q8

← JEE Main 2022 · 27 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 8 of 90

Let $f$ be a differentiable function in $\left(0,\dfrac\pi2\right)$ . If $\displaystyle\int_{\cos x}^1t^2f(t)\,dt=\sin^3x+\cos x$ , then $\dfrac1{\sqrt3}f'\left(\dfrac1{\sqrt3}\right)$ is equal to

6-9\sqrt2

6-\dfrac9{\sqrt2}

\dfrac92-6\sqrt2

\dfrac9{\sqrt2}-6

Answer and worked solution

Official answer: B

Differentiate. $\cos^2x\,f(\cos x)\sin x=3\sin^2x\cos x-\sin x$ , so with $t=\cos x$ : $f(t)=\dfrac{3\sqrt{1-t^2}}t-\dfrac1{t^2}$ .

Derivative. $f'(t)=-\dfrac3{t^2\sqrt{1-t^2}}+\dfrac2{t^3}$ . At $t=\frac1{\sqrt3}$ : $-\dfrac{9\sqrt3}{\sqrt2}+6\sqrt3$ .

Value. $\dfrac1{\sqrt3}f'=6-\dfrac9{\sqrt2}$ .

(B)

Topics:Leibniz rule Derivatives of standard functions

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 8