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If a ∫ 0 1 1 + 4 X 2 D X = π 8 , Then a Equals,π 2,1 2,π 4,1

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प्रश्न

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 

विकल्प

  • \[\frac{\pi}{2}\]
  • \[\frac{1}{2}\]
  • \[\frac{\pi}{4}\]
  • 1

MCQ
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उत्तर

\[\frac{1}{2}\]

\[\int_0^\alpha \frac{1}{1 + 4 x^2} d x = \frac{\pi}{8}\]

\[ \Rightarrow \int_0^\alpha \frac{1}{1 + \left( 2x \right)^2} d x = \frac{\pi}{8}\]

\[ \Rightarrow \frac{1}{2} \left[ \tan^{- 1} 2x \right]_0^\alpha = \frac{\pi}{8}\]

\[ \Rightarrow \frac{1}{2} \tan^{- 1} 2\alpha = \frac{\pi}{8}\]

\[ \Rightarrow 2\alpha = \tan\frac{\pi}{4}\]

\[ \Rightarrow 2\alpha = 1\]

\[ \therefore \alpha = \frac{1}{2}\]

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अध्याय 19: Definite Integrals - MCQ [पृष्ठ ११८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
MCQ | Q 21 | पृष्ठ ११८

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