If _0^ k d x 2+8 x^2 = 16 , then k = - Vidyadip

MCQ

If $\int_0^{ k } \frac{ d x}{2+8 x^2}=\frac{\pi}{16}$, then $k =$

A
1
✓
$\frac{1}{2}$
C
$\frac{1}{4}$
D
$\frac{1}{8}$

✓

Answer

Correct option: B.

$\frac{1}{2}$

(B)
$\int_0^{ k } \frac{ d x}{2+8 x^2}=\frac{\pi}{16}$
$\Rightarrow \frac{1}{2} \int_0^k \frac{1}{1^2+(2 x)^2} d x=\frac{\pi}{16}$
$\Rightarrow \frac{1}{2}\left[\frac{\tan ^{-1}(2 x)}{2}\right]_0^{ k }=\frac{\pi}{16}$
$\Rightarrow \frac{1}{4}\left(\tan ^{-1} 2 k \right)=\frac{\pi}{16}$
$\Rightarrow \tan ^{-1} 2 k =\frac{\pi}{4} \Rightarrow k =\frac{1}{2}$

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