If y= ^ -1 (1)/(x^2+x+1)+ ^ -1 (1)/(x^2+3 x+3)+ ^ -1 (1)/(x^2+5 x+7)+ .to n terms, then (d y)/(d x)=

Correct answer: B. (1)/((x+n)^2+1)-(1)/(x^2+1) Given, y= ^ -1 (1)/(x^2+x+1)+ ^ -1 (1)/(x^2+3 x+3)+ ^ -1 (1)/(x^2+5 x+7)+ · to n terms = ^ -1 \ (1)/(1+x(x+1))\ + ^ -1 \ (1)/(1+(x+1)(x+2))\ + ^ -1 \ (1)/(1+(x+2)(x+3))\ + + ^ -1 \ (1)/(1+(x+(n-1))(x+n))\ = ^ -1 \ ((x+1)-x)/(1+(x+1) x)\ + ^ -1 \ ((x+2)-(x+1))/(1+(x+2)(x+1))\ + ^ -1 \ ((x+3)-(x+2))/(1+(x+3)(x+2))\ + .+ ^ -1 (((x+n)-(x+n-1))/(1+(x+n)(x+n-1))) y=\ ^ -1 (x+1)- ^ -1 (x)\ +\ ^ -1 (x+2)- ^ -1 (x+1)\ +\ ^ -1 (x+3)- ^ -1 (x+2)\ + ..+\ ^ -1 (x+n)- ^ -1 [x+(n-1)]\ So, y= ^ -1 (x+n)- ^ -1 (x) On differentiating both sides w.r.t. x , we get (d y)/(d x)=(1)/(1+(x+n)^2)-(1)/(1+x^2)

BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Sequences and Series

BITSAT 2024 · Paper 1 · Mathematics · Sequences and Series

If y= ^ -1 (1)/(x^2+x+1)+ ^ -1 (1)/(x^2+3 x+3)+ ^ -1 (1)/(x^2+5...

y=\tan ^{-1} \frac{1}{x^2+x+1}+\tan ^{-1} \frac{1}{x^2+3 x+3}+\tan ^{-1} \frac{1}{x^2+5 x+7}+\ldots

.to

n

terms, then

\frac{d y}{d x}=

Options

A $\frac{1}{x^2+n^2}-\frac{1}{x^2+1}$
B $\frac{1}{(x+n)^2+1}-\frac{1}{x^2+1}$
C $\frac{1}{x^2+(n+1)^2}-\frac{1}{x^2+1}$
DNone of these

Correct answer

B. $\frac{1}{(x+n)^2+1}-\frac{1}{x^2+1}$

Step-by-step solution

Given,

y=\tan ^{-1} \frac{1}{x^2+x+1}+\tan ^{-1} \frac{1}{x^2+3 x+3}+\tan ^{-1} \frac{1}{x^2+5 x+7}+\cdots \cdot

n

terms

=\tan ^{-1}\left\{\frac{1}{1+x(x+1)}\right\}+\tan ^{-1}\left\{\frac{1}{1+(x+1)(x+2)}\right\}+\tan ^{-1}\left\{\frac{1}{1+(x+2)(x+3)}\right\}+\ldots \ldots \ldots+\tan ^{-1}\left\{\frac{1}{1+(x+(n-1))(x+n)}\right\}=\tan ^{-1}\left\{\frac{(x+1)-x}{1+(x+1) x}\right\}+\tan ^{-1}\left\{\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}\right\}+\tan ^{-1}\left\{\frac{(x+3)-(x+2)}{1+(x+3)(x+2)}\right\}+\ldots.+\tan ^{-1}\left(\frac{(x+n)-(x+n-1)}{1+(x+n)(x+n-1)}\right)\therefore y=\left\{\tan ^{-1}(x+1)-\tan ^{-1}(x)\right\}+\left\{\tan ^{-1}(x+2)-\tan ^{-1}(x+1)\right\}+\left\{\tan ^{-1}(x+3)-\tan ^{-1}(x+2)\right\}+\ldots \ldots..+\left\{\tan ^{-1}(x+n)-\tan ^{-1}[x+(n-1)]\right\}

So,

\quad y=\tan ^{-1}(x+n)-\tan ^{-1}(x)

On differentiating both sides w.r.t.

x

, we get

\frac{d y}{d x}=\frac{1}{1+(x+n)^2}-\frac{1}{1+x^2}

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