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Mathematics Question on Differential equations

Let f be a differentiable function in (0,π2)(0, \frac{π}{2}). If ∫cosx1t2f(t)dt=sin3x+cosx∫_{cosx} ^1 t^2f(t)dt=sin^3x+cosx, then 13f′(13)\frac{1}{\sqrt3}f'(\frac{1}{\sqrt3}) is equal to

A

6−926-9\sqrt2

B

6−926-\frac{9}{\sqrt2}

C

92−62\frac{9}{2}-6\sqrt2

D

92−6\frac{9}{\sqrt 2}-6

Answer

6−926-\frac{9}{\sqrt2}

Explanation

Solution

∫cosx1t2f(t)dt=sin3x+cosx∫_{cosx} ^1 t^2f(t)dt=sin^3x+cosx

⇒sin  x  cos2x  f(cosx)=3sin2x  cos  x–sin  x⇒ sin \;x \;cos^2 x\; f(cos x) = 3 sin^2 x \;cos \;x – sin \;x

⇒f(cos  x)=3  tanx–sec2x⇒ f(cos \;x) = 3 \;tan x – sec^2 x

⇒f′(cosx).(–sinx)=3sec2  x–2sec2  x  tan  x⇒ f′(cos x) . (– sin x) = 3 sec^2\; x – 2 sec^2 \;x \;tan \;x

Put cosx=13,cosx=\frac{1}{\sqrt3},

∴f′(13)(−23)=9−62∴f′\bigg(\frac{1}{\sqrt3}\bigg)\bigg(−\frac{\sqrt2}{\sqrt3}\bigg)=9−6\sqrt2

13f′(13)=6–92\frac{1}{\sqrt3}f′(\frac{1}{\sqrt3})=6–\frac{9}{\sqrt2}