Q106 / 155 · WBJEE 2019 · Mathematics
Let $I_{n}=\int_{0}^{1} x^{n} \tan ^{-1} x d x .$ If $a_{n} I_{n+2}+b_{n} I_{n}=c_{n}$ for all $n \geq 1,$ then
- A$a_{1}, a_{2}, a_{3}$ are in GP
- Correct answerB$b_{1}, b_{2}, b_{3}$ are in AP
- C$c_{1}, c_{2}, c_{3}$ are in HP
- Correct answerD$a_{1}, a_{2}, a_{3}$ are in $A P$
Solution
We have,
$I_{n}=\int_{0}^{1} x^{n} \tan ^{-1} x d x$
$=\left[\tan ^{-1} x \cdot \frac{x^{n+1}}{n+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{1}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1} \int_{0}^{1} \frac{x^{2} \cdot x^{n-1}}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1} \int_{0}^{1} \frac{\left(1+x^{2}\right) x^{n-1}-x^{n-1}}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1} \int_{0}^{1} x^{n-1} d x+\frac{1}{n+1} \int_{0}^{1} \frac{x^{n-1}}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1}\left[\frac{x^{n}}{n}\right]_{0}^{1}+\frac{1}{n+1}$
$\left[x^{n-1} \cdot \tan ^{-1} x\right]_{0}^{1}-\int_{0}^{1} \tan ^{-1} x \cdot(n-1) x^{n-2} d x$
$\begin{aligned}=\frac{\pi}{4(n+1)}-\frac{1}{n(n+1)}+\frac{\pi}{4(n+1)}-\frac{n-1}{n+1} & \int_{0}^{1} x^{n-2} \tan ^{-1} x d x \end{aligned}$
$\Rightarrow \quad I_{n}=\frac{\pi}{2(n+1)}-\frac{1}{n(n+1)}-\frac{n-1}{n+2} I_{n-2}$
$\Rightarrow \quad I_{n}+\frac{n-1}{n+2} I_{n-2}=\frac{\pi}{2(n+1)}-\frac{1}{n(n+1)}$
Put $n=n+2,$ we get $I_{n+2}+\frac{n+1}{n+3} I_{n}=\frac{\pi}{2(n+3)}-\frac{1}{(n+2)(n+3)}$
$\Rightarrow(n+3) I_{n+2}+(n+1) I_{n}=\frac{\pi}{2}-\frac{1}{n+2}$
$\therefore a_{n}=n+3, b_{n}=n+1, c_{n}=\frac{\pi}{2}-\frac{1}{n+2}$
$\therefore a_{n}$ and $b_{n}$ are in $\mathrm{AP}$.
Hence, the correct answer is options: 'B', 'D'.