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  • Please solve RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 21 maths textbook solution

Please solve RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 21 maths textbook solution

Answers (1)

Answer:

\frac{d y}{d x}=\{0, x \geq 0\} \text { does not exist for } x<0

Hint:

changing   \sec ^{-1}\left(\frac{x+1}{x-1}\right) \text { into } \cos ^{-1}\left(\frac{x-1}{x+1}\right)

Given:

y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x+1}{x-1}\right)
Solution:  

\begin{aligned} &y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x+1}{x-1}\right) \\\\ &y=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x+1}{x-1}\right) \end{aligned}

Which exists for

-1 \leq\left(\frac{x-1}{x+1}\right) \leq 1\text { and is equal to } \frac{\pi}{2}

Now,

\frac{x-1}{x+1} \leq 1

\begin{aligned} &\Rightarrow \frac{x-1}{x+1}-1 \leq 0 \\\\ &\Rightarrow \frac{x-1}{x+1}-\frac{x+1}{x+1} \leq 0 \\\\ &\Rightarrow-\frac{2}{x+1} \leq 0 \end{aligned}

\begin{aligned} &\Rightarrow x+1>0 \\\\ &\Rightarrow x>1 \ldots \ldots \ldots . . \text { (i) } \end{aligned}

Also,

\begin{aligned} &\frac{x-1}{x+1} \geq-1 \\\\ &\Rightarrow \frac{x-1}{x+1}+1 \geq 0 \end{aligned}

\begin{aligned} &\Rightarrow \frac{x-1}{x+1}+\frac{x+1}{x+1} \geq 0\\\\ &\Rightarrow \frac{2 x}{x+1} \geq 0\\\\ &\Rightarrow x \geq 0 \text { or } x<-1 \ldots \ldots \ldots \ldots \ldots \ldots(ii) \end{aligned}

Comparing equations (i) and (ii) we understand that the condition satisfying both inequalities is x\geq 0. So, for x\geq 0

y=\cos ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{x+1}{x-1}\right)=\frac{\pi}{2}, which is a constant

So,

\frac{d y}{d x}=\{0, x \geq 0\} \text { does not exist for } x<0

Posted by

infoexpert26

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