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Question: If \(F(x) = \frac{1}{x^{2}}\int_{4}^{x}{(4t^{2} - 2F^{'}(t))dt,}\) then \(F^{'}(4)\) equals...

If F(x)=1x2∫4x(4t2−2F′(t))dt,F(x) = \frac{1}{x^{2}}\int_{4}^{x}{(4t^{2} - 2F^{'}(t))dt,} then F′(4)F^{'}(4) equals

A

32

B

323\frac{32}{3}

C

329\frac{32}{9}

D

None of these

Answer

329\frac{32}{9}

Explanation

Solution

We have F(x)=1x2∫4x(4t2−2F′(t))dtF(x) = \frac{1}{x^{2}}\int_{4}^{x}{(4t^{2} - 2F'(t))dt}

∴F′(x)=1x2(4x2−2F′(x))−2x3∫4x(4t2−2F′(t))dt\therefore F'(x) = \frac{1}{x^{2}}\left( 4x^{2} - 2F'(x) \right) - \frac{2}{x^{3}}\int_{4}^{x}{(4t^{2} - 2F'(t))dt}

⇒ F′(4)=116[64−2F′(4)]−0⇒F′(4)=329F'(4) = \frac{1}{16}\lbrack 64 - 2F'(4)\rbrack - 0 \Rightarrow F'(4) = \frac{32}{9}.