Let f:R→R be a differentiable function such that fπ4=2, fπ2=0 and f′π2=1 and let g(x)=∫xπ/4 f′(t)sec⁡t+tan⁡sec tf(t)dt for x∈π4,π2. Then limx→π2− g(x) is equal to :

g(x)=∫xπ/4 f′(t)sec⁡t+tan⁡tsec⁡tf(t)dtg(x)=∫xπ/4 d(f(t)⋅sec⁡t)=f(t)sec⁡txπ/4 g(x)=2−f(x)sec⁡x=2−f(x)cos⁡xlim x→π2- g(x)=2−limx→π2− f(x)cos⁡x=2−limx→π2− f′(x)(−sin⁡x) by L-Hospital rule=2+f′π2sin⁡π2=2+11=3

Let f:R→R be a differentiable function such that fπ4=2, fπ2=0 and f′π2=1 and let g(x)=∫xπ/4 f′(t)sec⁡t+tan⁡sec tf(t)dt for x∈π4,π2. Then limx→π2− g(x) is equal to :

g(x)=∫xπ/4 f′(t)sec⁡t+tan⁡tsec⁡tf(t)dtg(x)=∫xπ/4 d(f(t)⋅sec⁡t)=f(t)sec⁡txπ/4 g(x)=2−f(x)sec⁡x=2−f(x)cos⁡xlim x→π2- g(x)=2−limx→π2− f(x)cos⁡x=2−limx→π2− f′(x)(−sin⁡x) by L-Hospital rule=2+f′π2sin⁡π2=2+11=3

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Let $f : R \to R$ be a differentiable function such that $f (\frac{π}{4}) = \sqrt{2}, f (\frac{π}{2}) = 0 and f^{'} (\frac{π}{2}) = 1$ and let $g (x) = \int_{x}^{π / 4} (f^{'} (t) \sec t + \tan \sec tf (t)) dt$ for $x \in [\frac{π}{4}, \frac{π}{2})$ . Then $lim_{x \to {(\frac{π}{2})}^{-}} g (x)$ is equal to :

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Detailed Solution

$\begin{array}{l} g (x) = \int_{x}^{π / 4} (f^{'} (t) \sec t + \tan tsec tf (t)) dt \\ g (x) = \int_{x}^{π / 4} d (f (t) \cdot \sec t) = {f (t) \sec t|}_{x}^{π / 4} \\ g (x) = 2 - f (x) \sec x = 2 - (\frac{f (x)}{\cos x}) \\ \lim_{x \to {(\frac{π}{2})}^{-}} g (x) = 2 - lim_{x \to {(\frac{π}{2})}^{-}} (\frac{f (x)}{\cos x}) \end{array}$

$\begin{array}{l} = 2 - lim_{x \to {(\frac{π}{2})}^{-}} \frac{f^{'} (x)}{(- \sin x)} (b y L - H o s p i t a l r u l e) \\ = 2 + \frac{f^{'} (\frac{π}{2})}{\sin \frac{π}{2}} = 2 + \frac{1}{1} = 3 \end{array}$