Question

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 G.P
(B) \(b_{1}, b_{2}, b_{3}\) are in A.P
(C) \(\mathrm{c}_{1}, \mathrm{c}_{2}, \mathrm{C}_{3}\) are in H.P
(D) \(a_{1}, a_{2}, a_{3}\) are in A.P Ans: (B,D)

Answer 1

Ans: (B,D)
Hint \(:\left.(n+1)\right|_{n}+\left.(n+3)\right|_{n+2}=\int_{0}^{1}\left\{(n+1) x^{n}+(n+3) x^{n+2}\right\} \tan ^{-1} x d x\)
\(=\left[\left\{(n+1) \frac{x^{n+1}}{n+1}+(n+3) \frac{x^{n+3}}{n+3}\right\} \tan ^{-1} x\right]_{0}^{1}-\int_{0}^{1} x^{n+1}\left(1+x^{2}\right) \cdot \frac{1}{1+x^{2}} d x=\left[\left(x^{n+1}+x^{n+3}\right) \tan ^{-1} x\right]_{0}^{1}-\left[\frac{x^{n+2}}{n+2}\right]_{0}^{1}\)
\(=\frac{\pi}{2}-\frac{1}{n+2}\)
\(c_{n}=\frac{\pi}{2}-\frac{1}{n+2}\)
\(\therefore a_{n}=n+3 ; b_{n}=n+1 ; c_{n}=\frac{\pi}{2}-\frac{1}{n+2}\)

Answer 2

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Answer 3

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