Skip to main content

Updates ...

Visit the website 123iitjee.manishverma.site for latest posts, courses, admission & more.

For guest/sponsored article(s), please check this link.

$f(x) = {\log _2}\left( {1 + \tan \left( {\frac{{\pi x}}{4}} \right)} \right)$

$\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ............ + f(1)} \right)$=?

Let $f: (0, 2) \to R $ be defined as $f(x) = {\log _2}\left( {1 + \tan \left( {\frac{{\pi x}}{4}} \right)} \right)$. Then $\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ............ + f(1)} \right)$ is equal to _ _ _ _ .

Solution

The given limit $\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ............ + f(1)} \right)$

$ = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{r = 1}^n {f\left( {\frac{r}{n}} \right)} } \right).\frac{2}{n}$

$ = \int\limits_0^1 {f(x).2dx = I} $

$I = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \tan \frac{{\pi x}}{4}} \right)} dx$

$ \Rightarrow I = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \tan \frac{{\pi (1 - x)}}{4}} \right)} dx = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \tan \left( {\frac{\pi }{4} - \frac{{\pi x}}{4}} \right)} \right)} dx$

$ \Rightarrow I = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \frac{{1 - \tan \frac{{\pi x}}{4}}}{{1 + \tan \frac{{\pi x}}{4}}}} \right)} dx = 2\int\limits_0^1 {{{\log }_2}\left( {\frac{2}{{1 + \tan \frac{{\pi x}}{4}}}} \right)} dx$

$ \Rightarrow I = 2\int\limits_0^1 {\left[ {\log {}_22 - {{\log }_2}\left( {1 + \tan \frac{{\pi x}}{4}} \right)} \right]} dx = 2\int\limits_0^1 {\left[ {1 - {{\log }_2}\left( {1 + \tan \frac{{\pi x}}{4}} \right)} \right]} dx$

$ \Rightarrow I = 2 - I$

$ \Rightarrow I = 1$

Popular posts from this blog

$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$ $\Rightarrow n=7$