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

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Definite Integrals

Let

f : (0, 2) \to R

be defined as

f (x) = lo g_{2} (1 + tan (\frac{π x}{4}))

. Then,

lim_{n \to \infty} \frac{2}{n} (f (\frac{1}{n}) + f (\frac{2}{n}) + \dots . + f (1))

is equal to

Answer: 1

The bracket is a Riemann sum, so

n \to \infty lim \frac{2}{n} r = 1 \sum n f (\frac{r}{n}) = 2 \int_{0}^{1} lo g_{2} (1 + tan \frac{π x}{4}) d x

Write

J

for that integral and replace

x

1 - x

. Since

tan (\frac{π}{4} - \frac{π x}{4}) = \frac{1 - tan \frac{π x}{4}}{1 + tan \frac{π x}{4}}, 1 + tan (\frac{π}{4} - \frac{π x}{4}) = \frac{2}{1 + tan \frac{π x}{4}}

the substitution turns

J

into

J = \int_{0}^{1} lo g_{2} \frac{2}{1 + tan \frac{π x}{4}} d x = \int_{0}^{1} [1 - lo g_{2} (1 + tan \frac{π x}{4})] d x = 1 - J

Hence

J = \frac{1}{2}

, and the required limit is

2 J = 1

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer