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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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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$