If -0xft dt-x2-x1t2ft dt- then f-12 is-

∫_0^xf(t) dt=x^2+∫_x^1t^2f(t) dt Differentiating w.r.t x f(x) 0=2x+0 x^2f(x) ⇒ f(x)=2x1+x^2 ⇒ nbsp; f'(x)=(1+x^2)2 2x·2x(1+x^2)^2 ⇒ f'(x)=2 2x^2(1+x^2)^2 Henc ...

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Standard XI

Mathematics

Differentiation under Integral Sign

If ∫0xft dt=x...

Question

If $x \int 0 f (t) d t = x^{2} + 1 \int x t^{2} f (t) d t$ , then $f^{'} (\frac{1}{2})$ is:

\frac{18}{25}

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\frac{4}{5}

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\frac{24}{25}

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\frac{6}{25}

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Solution

The correct option is C $\frac{24}{25}$
$x \int 0 f (t) d t = x^{2} + 1 \int x t^{2} f (t) d t$
Differentiating w.r.t $x$
$f (x) - 0 = 2 x + 0 - x^{2} f (x)$
$\Rightarrow f (x) = \frac{2 x}{1 + x^{2}}$
$\Rightarrow f^{'} (x) = \frac{(1 + x^{2}) 2 - 2 x \cdot 2 x}{(1 + x^{2})^{2}}$
$\Rightarrow f^{'} (x) = \frac{2 - 2 x^{2}}{(1 + x^{2})^{2}}$
Hence, $f^{'} (\frac{1}{2}) = \frac{24}{25}$

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If x∫0f(t) dt=x2+1∫xt2f(t) dt, then f′(12) is:

The correct option is C 2425x∫0f(t) dt=x2+1∫xt2f(t) dt Differentiating w.r.t x f(x)−0=2x+0−x2f(x) ⇒f(x)=2x1+x2 ⇒f′(x)=(1+x2)2−2x⋅2x(1+x2)2 ⇒f′(x)=2−2x2(1+x2)2 Hence, f′(12)=2425

If $x \int 0 f (t) d t = x^{2} + 1 \int x t^{2} f (t) d t$ , then $f^{'} (\frac{1}{2})$ is: