Question

Let f be a differentiable function in \((0, \frac{π}{2})\). If \(∫_{cosx} ^1 t^2f(t)dt=sin^3x+cosx\), then \(\frac{1}{\sqrt3}f'(\frac{1}{\sqrt3})\) is equal to

• A. \(6-9\sqrt2\)
• B. \(6-\frac{9}{\sqrt2}\)
• C. \(\frac{9}{2}-6\sqrt2\)
• D. \(\frac{9}{\sqrt 2}-6\)

Answer 1

The Correct Option is B

\(∫_{cosx} ^1 t^2f(t)dt=sin^3x+cosx\)

\(⇒ sin \;x \;cos^2 x\; f(cos x) = 3 sin^2 x \;cos \;x – sin \;x\)

\(⇒ f(cos \;x) = 3 \;tan x – sec^2 x\)

\(⇒ f′(cos x) . (– sin x) = 3 sec^2\; x – 2 sec^2 \;x \;tan \;x\)

Put \(cosx=\frac{1}{\sqrt3},\)

\(∴f′\bigg(\frac{1}{\sqrt3}\bigg)\bigg(−\frac{\sqrt2}{\sqrt3}\bigg)=9−6\sqrt2\)

\(\frac{1}{\sqrt3}f′(\frac{1}{\sqrt3})=6–\frac{9}{\sqrt2}\)