MCQ
જો $\int_0^k {\frac{{dx}}{{2 + 8{x^2}}}} = \frac{\pi }{{16}}\,,$ તો $k = $
- A$1$
- ✓$\frac{1}{2}$
- C$\frac{1}{4}$
- Dએકપણ નહીં.
$ = \frac{1}{4}|{\tan ^{ - 1}}t|_0^{2k} = \frac{1}{4}{\tan ^{ - 1}}2k$.
Comparing it with the given value, we get
${\tan ^{ - 1}}2k = \frac{\pi }{4} $
$\Rightarrow 2k = 1$
$ \Rightarrow k = \frac{1}{2}$.
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