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JEE Mains · Maths · STD 12 - 7.2 definite integral

ધારો કે \(f: R \rightarrow R\) વિકલનીય વિધેય છે કે જેથી \(f\left(\frac{\pi}{4}\right)=\sqrt{2}, f\left(\frac{\pi}{2}\right)=0\) અને \(f^{\prime}\left(\frac{\pi}{2}\right)=1\) તથા ધારો કે \(x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\) માટે \(g(x)=\int_{x}^{\pi / 4}\left(f^{\prime}( t ) \operatorname{sect}+\operatorname{tant} \operatorname{sect} f( t )\right) dt\), તો \(\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g(x)=\)...........

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(-3\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\(g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t\) \(g(x)=\int\limits_{x}^{\pi / 4} d ( f ( t ) \cdot \sec t )=\left. f ( t ) \operatorname{sect}\right|_{ x } ^{\pi / 4}\)…
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