JEE Main 2021 (Online) 16th March Morning ShiftQuestion 1
Let $f:(0,2)\to\mathbf{R}$ be defined as $f(x)=\log_2\left(1+\tan\left(\dfrac{\pi x}{4}\right)\right)$.
Then, $\displaystyle\lim_{n\to\infty}\dfrac{2}{n}\left(f\left(\dfrac{1}{n}\right)+f\left(\dfrac{2}{n}\right)+\ldots+f(1)\right)$ is equal to $\_\_\_\_\_\_\_\_$ .
Definite Integration
Answer$1$
Solution
Concept. The limit is $2\int_0^1f$. With $\theta=\frac{\pi x}4$ it becomes the classic $\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta$. Formulas used. (1) $\lim\frac1n\sum_{k=1}^nf\big(\frac kn\big)=\int_0^1f$(2) $1+\tan\big(\frac\pi4-\theta\big)=\dfrac{2}{1+\tan\theta}$, which gives $\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta=\frac\pi8\ln2$Step 1 - Use (1). The limit is $2\int_0^1\log_2\big(1+\tan\frac{\pi x}4\big)dx$. Step 2 - Put $\theta=\frac{\pi x}4$. $dx=\frac4\pi d\theta$, $\theta:0\to\frac\pi4$: $$\int_0^1\ln\Big(1+\tan\frac{\pi x}4\Big)dx=\frac4\pi\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta.$$ Step 3 - Use (2). King's rule gives $J=\int_0^{\pi/4}\big(\ln2-\ln(1+\tan\theta)\big)d\theta=\frac\pi4\ln2-J$, so $J=\frac\pi8\ln2$. Hence the integral in Step 2 is $\frac12\ln2$. Step 4 - Value. $2\cdot\frac1{\ln2}\cdot\frac{\ln2}2=1$. Answer: 1