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Solve ∫ t²f(t)dt = 1−sinx for f(1/√3) | JEE

JEE Maths question with a full step-by-step solution.

Question
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is
A13\dfrac{1}{\sqrt3}
B3\sqrt3
C13\dfrac13
D33correct
Solution
Step 1: Differentiate both sides with respect to xx. The upper limit is constant, the lower limit is sin⁡x\sin x, so by Leibniz's rule
ddx∫sin⁡x1t2f(t) dt=−(sin⁡x)2f(sin⁡x)⋅cos⁡x.\dfrac{d}{dx}\int_{\sin x}^{1}t^{2}f(t)\,dt=-\big(\sin x\big)^{2}f(\sin x)\cdot\cos x.
 Step 2: Differentiate the right-hand side:
ddx(1−sin⁡x)=−cos⁡x.\dfrac{d}{dx}\big(1-\sin x\big)=-\cos x.
 Step 3: Equate the two derivatives:
−sin⁡2x f(sin⁡x)cos⁡x=−cos⁡x ⇒ sin⁡2x f(sin⁡x)=1 ⇒ f(sin⁡x)=1sin⁡2x.-\sin^{2}x\,f(\sin x)\cos x=-\cos x\ \Rightarrow\ \sin^{2}x\,f(\sin x)=1\ \Rightarrow\ f(\sin x)=\dfrac{1}{\sin^{2}x}.
 Step 4: Set sin⁡x=13\sin x=\dfrac{1}{\sqrt3}:
f ⁣(13)=1(1/3)2=11/3=3.f\!\left(\dfrac{1}{\sqrt3}\right)=\dfrac{1}{\big(1/\sqrt3\big)^{2}}=\dfrac{1}{1/3}=3.
 Correct answer: (4)
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