Skip to main content

Visit this link for 1 : 1 LIVE Classes.

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

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

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

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

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)