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$I = \int\limits_0^1 {\frac{{\ln (1 + x + {x^2} + .........\infty \, terms)}}{x}dx} $

A) $I = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ........\infty \, terms$

B) $I = 1 + \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + ........\infty \, terms$

C) $I = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ........\infty \, terms$

D) $I = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ........\infty \, terms$

Since x lies between 0 and 1, the infinite terms summation formula of an infinite G.P. with x as common ratio can be used here.

$I = \int\limits_0^1 {\frac{{\ln \left( {\frac{1}{{1 - x}}} \right)}}{x}dx} $

$ = \int\limits_0^1 {\frac{{ - \ln (1 - x)}}{x}dx} $

$ = \int\limits_0^1 {\frac{{x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \frac{{{x^4}}}{4} + .......}}{x}dx} $

$ = \int\limits_0^1 {1 + \frac{x}{2} + \frac{{{x^2}}}{3} + \frac{{{x^3}}}{4} + .......dx} $

$ = \left. {x + \frac{{{x^2}}}{{{2^2}}} + \frac{{{x^3}}}{{{3^2}}} + \frac{{{x^4}}}{{{4^2}}} + .........} \right|_0^1$

$ = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + .........$

Hence, option (C).

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The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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