If -0x ft d t-x2-x1 t2 ft d t- then f-1-2A- 4-5B- 24-25C- 18-25D- 6-25

∫_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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiation under Integral Sign

If ∫0x ft d t...

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{24}{25}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{6}{25}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $f^{'} (\frac{1}{2}) = \frac{24}{25}$

Suggest Corrections

Similar questions

Q. The vertex of the conic represented by

25 (x^{2} + y^{2}) = (3 x - 4 y + 12)^{2}

If in a triangle ABC, a=6cm, b=8cm, c=10cm, then the value of sin 2A is

\int_{0}^{\infty} \frac{d x}{(x + \sqrt{x^{2} + 1})^{5}} =

Factors of

$- x^{3} - 125$

Q. Let

f

be a non-negative function in

[0, 1]

and twice differentiable in

(0, 1) .

x \int 0 \sqrt{1 - (f^{'} (t))^{2}} d t = x \int 0 f (t) d t, 0 \leq x \leq 1

and

f (0) = 0,

then

lim x \to 0 \frac{1}{x^{2}} x \int 0 f (t) d t

Join BYJU'S Learning Program

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: