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if \int_{sinx}^{1}t^{2}f(t)dt=1-sinx,\: \; x\epsilon \left ( 0,\frac{\Pi }{2} \right )\; then f\left ( \frac{1}{\sqrt{3}} \right ) =

  • Option 1)

    3

  • Option 2)

    \frac{1}{3}

  • Option 3)

    \frac{1}{\sqrt3}

  • Option 4)

    \sqrt3

 

Answers (1)

best_answer

As we learnt

NEWTON LEIBNITZ THEOREM -

\frac {d}{dt}\left ( \int_{f(t)}^{\phi (t))}F(x)dx \right )=F(\phi(t))\phi^{'}(t)-F(f(t))f^{'}(t)

-

 

 

On differentiating both sides, we get

                         \Rightarrow \: -\sin^{2}xf\left ( \sin x \right )\cos x= -\cos x

                     \Rightarrow f\left ( \sin x \right )= cosec^{2}\, x

                       \Rightarrow f\left ( x \right )=\frac{1}{x^{2}}

                     \Rightarrow f\left ( \frac{1}{\sqrt{3}} \right )=3 

 


Option 1)

3

Option 2)

\frac{1}{3}

Option 3)

\frac{1}{\sqrt3}

Option 4)

\sqrt3

Posted by

gaurav

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