Limit Continuity and Differentiability 7 Question 5
5. Let $K$ be the set of all real values of $x$, where the function $f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ is not differentiable. Then, the set $K$ is equal to
(2019 Main, 11 Jan II)
(a) ${0}$
(b) $\varphi$ (an empty set)
(c) ${\pi}$
(d) ${0, \pi}$
where, $a$ and $b$ are non-negative real numbers. Determine the compositie function gof. If $(g \circ f)(x)$ is continuous for all real $x$ determine the values of $a$ and $b$. Further, for these values of $a$ and $b$, is gof differentiable at $x=0$ ? Justify your answer.
$(2002,5$ M)
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Answer:
Correct Answer: 5. (a)
Solution:
- Given expression is
$$ 2 y=\cot ^{-1} \frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}=\cot ^{-1} \frac{\sqrt{3} \cot x+1}{\cot x-\sqrt{3}} $$
[dividing each term of numerator and denominator by $\sin x]$
$$ \begin{aligned} & =\cot ^{-1} \frac{\cot \frac{\pi}{6} \cot x+1}{\cot x-\cot \frac{\pi}{6}} \\ & =\cot ^{-1} \cot \frac{\pi}{6}-x \quad{ }^{2} \quad \because \cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A} \end{aligned} $$
$$ \begin{aligned} & =\quad \frac{\pi}{6}-x^{2}, \quad 0<x<\frac{\pi}{6} \\ & \pi+\frac{\pi}{6}-x \quad, \quad \frac{\pi}{6}<x<\frac{\pi}{2} \\ & \because \cot ^{-1}(\cot \theta)=\begin{array}{cc} \pi+\theta, & -\pi<\theta<0 \\ \theta, & 0<\theta<\pi \\ \theta-\pi, & \pi<\theta<2 \pi \end{array} \\ & \Rightarrow 2 y=\begin{array}{l} \frac{\pi}{6}-x^{2}, \quad 0<x<\frac{\pi}{6} \\ \frac{7 \pi}{6}-x^{2}, \frac{\pi}{6}<x<\frac{\pi}{2} \end{array} \\ & \Rightarrow 2 \frac{d y}{d x}=\begin{array}{ll} 2 \frac{\pi}{6}-x(-1), & 0<x<\frac{\pi}{6} \\ 2 \frac{7 \pi}{6}-x(-1), & \frac{\pi}{6}<x<\frac{\pi}{2} \end{array} \\ & \Rightarrow \frac{d y}{d x}=\begin{array}{l} x-\frac{\pi}{6}, \quad 0<x<\frac{\pi}{6} \\ x-\frac{7 \pi}{6}, \quad \frac{\pi}{6}<x<\frac{\pi}{2} \end{array} \end{aligned} $$
