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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
If $\int_0^k {\frac{{dx}}{{2 + 8{x^2}}}} = \frac{\pi }{{16}}\,,$ then $k = $
  • A
    $1$
  • ✓
    $\frac{1}{2}$
  • C
    $\frac{1}{4}$
  • D
    None of these
✓

Answer

Correct option: B.
$\frac{1}{2}$
b
(b) $\int_0^k {\frac{1}{{2 + 8{x^2}}}dx = \frac{1}{2}\int_0^k {\frac{{dx}}{{1 + {{(2x)}^2}}}= \frac{1}{4}\int_0^{2k} {\frac{{dt}}{{1 + {t^2}}}} } } $

$ = \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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