यदि $f(1) = 1$ और $f'(1) = 2$ है,तो $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ का मान ज्ञात कीजिए।

  • A
    $2$
  • B
    $4$
  • C
    $1$
  • D
    $1/2$

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मान लीजिए $f : R \rightarrow R$ एक अवकलनीय फलन है,इस प्रकार कि $f \left(\frac{\pi}{4}\right)=\sqrt{2}$,$f \left(\frac{\pi}{2}\right)=0$ और $f^{\prime}\left(\frac{\pi}{2}\right)=1$ है। यदि $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t$ जहाँ $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$ है,तो $\lim\limits _{ x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g ( x )$ का मान ज्ञात कीजिए।

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