Limit Continuity And Differentiability Ques 45
- If $2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^2, x \in\left(0, \frac{\pi}{2}\right)$ then $\frac{d y}{d x}$ is equal to
(2019 Main, 8 April 1)
(a) $\frac{\pi}{6}-x$
(b) $x-\frac{\pi}{6}$
(c) $\frac{\pi}{3}-x$
(d) $2 x-\frac{\pi}{3}$
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Answer:
Correct Answer: 45.(b)
Solution: (b) Given expression is
$2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^2=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cot x+1}{\cot x-\sqrt{3}}\right)\right)^2$
[dividing each term of numerator and denominator by $\sin x$
$=\left(\cot ^{-1}\left(\frac{\cot \frac{\pi}{6} \cot x+1}{\cot x-\cot \frac{\pi}{6}}\right)\right)^2 \quad\left[\because \cot \frac{\pi}{6}=\sqrt{3}\right]$
$=\left(\cot ^{-1}\left(\cot \left(\frac{\pi}{6}-x\right)\right)\right)^2 \quad \left[ \because \ \cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A}\right]$
$ \begin{aligned} & =\left\{\begin{array}{cc} \left(\frac{\pi}{6}-x\right)^2, & 0<x<\frac{\pi}{6} \\ \left(\pi+\left(\frac{\pi}{6}-x\right)\right)^2, & \frac{\pi}{6}<x<\frac{\pi}{2} \end{array}\right. \\ & {\left[\because \cot ^{-1}(\cot \theta)=\left\{\begin{array}{cc} \pi+\theta, & -\pi<\theta<0 \\ \theta, & 0<\theta<\pi \\ \theta-\pi, & \pi<\theta<2 \pi \end{array}\right]\right.} \\ & \Rightarrow 2 y= \begin{cases}\left(\frac{\pi}{6}-x\right)^2, & 0<x<\frac{\pi}{6} \\ \left(\frac{7 \pi}{6}-x\right)^2, & \frac{\pi}{6}<x<\frac{\pi}{2}\end{cases} \\ & \Rightarrow 2 \frac{d y}{d x}= \begin{cases}2\left(\frac{\pi}{6}-x\right)(-1), & 0<x<\frac{\pi}{6} \\ 2\left(\frac{7 \pi}{6}-x\right)(-1), & \frac{\pi}{6}<x<\frac{\pi}{2}\end{cases} \\ & \Rightarrow \frac{d y}{d x}= \begin{cases}x-\frac{\pi}{6}, & 0<x<\frac{\pi}{6} \\ x-\frac{7 \pi}{6}, & \frac{\pi}{6}<x<\frac{\pi}{2}\end{cases} \\ \end{aligned} $
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