Question

Consider the function $f\left( x \right)=\left| \sin x \right|+\left| \cos x \right|$ for $0 < x < 2\pi$. Then
(a) $f\left( x \right)$is differentiable $\forall x\in \left( 0,2\pi \right)$
(b) $f\left( x \right)$is not differentiable at $x=\dfrac{\pi }{2},\pi$and $\dfrac{3\pi }{2}$and differentiable at all other values in $\left( 0,2\pi \right)$
(c) $f\left( x \right)$is not differentiable at $x=\dfrac{\pi }{2}$and $\dfrac{3\pi }{2}$and differentiable at all other values in $\left( 0,2\pi \right)$
(d) $f$is discontinuous at $x=\dfrac{\pi }{2},\pi$and $\dfrac{3\pi }{2}$

Answer 1

Hint: A function is differentiable at $x=a$ , if the left-hand derivative of the function is equal to the right-hand derivative of the function at $x=a$ .

Complete step-by-step answer:
We know \left| \sin x \right|=\left\\{ \begin{aligned} & \sin x,0\le x\le \pi \\\ & -\sin x,\pi \le x\le 2\pi \\\ \end{aligned} \right.
and \left| \cos x \right|=\left\\{ \begin{aligned} & \cos x,\text{ }0\le x\le \dfrac{\pi }{2} \\\ & -\cos x,\text{ }\dfrac{\pi }{2}\le x\le \dfrac{3\pi }{2} \\\ & \cos x,\text{ }\dfrac{3\pi }{2}\le x\le 2\pi \\\ \end{aligned} \right.
So, we can rewrite the function as,

& \sin x+\cos x,\text{ }0\le x\le \dfrac{\pi }{2} \\\ & \sin x-\cos x,\text{ }\dfrac{\pi }{2}\le x\le \pi \\\ & -\sin x-\cos x,\text{ }\pi \le x\le \dfrac{3\pi }{2} \\\ & -\sin x+\cos x,\text{ }\dfrac{3\pi }{2}\le x\le 2\pi \\\ \end{aligned} \right.$$ Now, for $$f\left( x \right)$$ to be differentiable at $$x=a$$ , the left-hand derivative should be equal to the right-hand derivative at $$x=a$$ . We know, the left-hand derivative of $$f\left( x \right)$$ at $$x=a$$ is given as $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}$$ and the right-hand derivative of $$f\left( x \right)$$ at $$x=a$$ is given as $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$$ . Now, we will check the differentiability of $$f\left( x \right)$$ at $$\dfrac{\pi }{2}$$ . The left-hand derivative of $$f\left( x \right)$$ at $$x=\dfrac{\pi }{2}$$ is given as $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( \dfrac{\pi }{2}-h \right)-f\left( \dfrac{\pi }{2} \right)}{-h}$$ . $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{\pi }{2}-h \right)+\cos \left( \dfrac{\pi }{2}-h \right)-\left( \sin \dfrac{\pi }{2}+\cos \dfrac{\pi }{2} \right)}{-h}$$ $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\cos \text{ }h+\sin \text{ }h-1}{-h}$$ Now, on substituting $$h=0$$ in the limit, we can see that it gives an indeterminate value $$\dfrac{0}{0}$$ . In such conditions, we apply L’ Hopital’s rule to evaluate the limit. L’ Hopital’s Rule states that “ if $$L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}$$ and $$f(x)=g(x)=0$$ or $$\infty $$ , then $$L=\underset{x\to c}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}$$ .” So, to find the value of the limit, we must differentiate the numerator and the denominator with respect to $$x$$ . So, $$L'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{-sin }h+\cos \text{ }h}{-1}$$ $$=-1$$ The right-hand derivative of $$f\left( x \right)$$ at $$x=\dfrac{\pi }{2}$$ is given as $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( \dfrac{\pi }{2}+h \right)-f\left( \dfrac{\pi }{2} \right)}{h}$$ . $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{\pi }{2}+h \right)-\cos \left( \dfrac{\pi }{2}+h \right)-\left( \sin \dfrac{\pi }{2}-\cos \dfrac{\pi }{2} \right)}{h}$$ $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\cos \text{ }h+\sin \text{ }h-1}{h}$$ Again, applying L’ Hopital’s rule, we get $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{-sin }h+\cos \text{ }h}{1}$$ . $$=1$$ The left-hand derivative of $$f\left( x \right)$$ is not equal to the right-hand derivative of $$f\left( x \right)$$ at $$x=\dfrac{\pi }{2}$$ . So, the function is not differentiable at $$x=\dfrac{\pi }{2}$$ . Now, we will check the differentiability of $$f\left( x \right)$$ at $$x=\pi $$ . The left-hand derivative of $$f\left( x \right)$$ at $$x=\pi $$ is given as $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( \pi -h \right)-f\left( \pi \right)}{-h}$$ . $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sin \left( \pi -h \right)-\cos \left( \pi -h \right)-\left( \sin \pi -\cos \pi \right)}{-h}$$ $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\cos \text{ }h+\sin \text{ }h-1}{-h}$$ Applying L’ Hopital’s rule, we get $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{-sin }h+\cos \text{ }h}{-1}$$ . $$=\dfrac{0+1}{-1}=-1$$ The right-hand derivative of $$f\left( x \right)$$ at $$x=\pi $$ is given as $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( \pi +h \right)-f\left( \pi \right)}{h}$$ . $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sin \left( \pi +h \right)-\cos \left( \pi +h \right)-\left( -\sin \pi -\cos \pi \right)}{h}$$ $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{sin }h+\cos \text{ }h-1}{h}$$ Applying L’ Hopital’s rule, we get $${{R}^{'}}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\text{cos }h-\sin \text{ }h}{1}$$ . $$=1$$ The left-hand derivative is not equal to the right-hand derivative. So, the function $$f\left( x \right)$$ is not differentiable at $$x=\pi $$ . Now, we will check the differentiability of $$f\left( x \right)$$ at $$x=\dfrac{3\pi }{2}$$ . The left-hand derivative of $$f\left( x \right)$$ at $$x=\dfrac{3\pi }{2}$$ is given as $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( \dfrac{3\pi }{2}-h \right)-f\left( \dfrac{3\pi }{2} \right)}{-h}$$ . $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sin \left( \dfrac{3\pi }{2}-h \right)-\cos \left( \dfrac{3\pi }{2}-h \right)-\left( -\sin \dfrac{3\pi }{2}-\cos \dfrac{3\pi }{2} \right)}{-h}$$ $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{cos }h+\sin \text{ }h-\left( 1 \right)}{-h}$$ Applying L’ Hopital’s rule, we get $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{-sin }h+\cos \text{ }h}{-1}=-1$$ . The right-hand derivative of $$f\left( x \right)$$ at $$x=\dfrac{3\pi }{2}$$ is given as $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( \dfrac{3\pi }{2}+h \right)-f\left( \dfrac{3\pi }{2} \right)}{h}$$ . $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{-\sin \left( \dfrac{3\pi }{2}+h \right)+\cos \left( \dfrac{3\pi }{2}+h \right)-\left( -\sin \dfrac{3\pi }{2}+\cos \dfrac{3\pi }{2} \right)}{h}$$ $$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{cos }h+\sin \text{ }h-\left( 1 \right)}{h}$$ Applying L’ Hopital’s rule, we get $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\text{-sin }h+\cos \text{ }h}{1}$$ . $$=1$$ The left-hand derivative of $$f\left( x \right)$$ is not equal to the right-hand derivative of $$f\left( x \right)$$ at $$x=\dfrac{3\pi }{2}$$ . So, the function is not differentiable at $$x=\dfrac{3\pi }{2}$$ . Now, we will check the continuity of the function at critical points. A function is continuous at a point if the value of limit at that point is equal to the value of the function at that point. The left-hand limit of the function at $\dfrac{\pi }{2}$ is given as: $$\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\left( \sin x+\cos x \right)=\left( \sin \dfrac{\pi }{2}+\cos \dfrac{\pi }{2} \right)=1$$ The right-hand limit of the function at $\dfrac{\pi }{2}$ is given as: $$\underset{x\to {{\dfrac{\pi }{2}}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{\dfrac{\pi }{2}}^{+}}}{\mathop{\lim }}\,\left( \sin x-\cos x \right)=\left( \sin \dfrac{\pi }{2}-\cos \dfrac{\pi }{2} \right)=1$$ The value of the function at $\dfrac{\pi }{2}$ is given as $f\left( \dfrac{\pi }{2} \right)=\sin \dfrac{\pi }{2}+\cos \dfrac{\pi }{2}=1$ . We can see that the value of the left-hand limit of the function is equal to the value of the right-hand limit of the function at $\dfrac{\pi }{2}$ and both are equal to the value of the function at $\dfrac{\pi }{2}$ . So, the function is continuous at $\dfrac{\pi }{2}$ . So, option D. is incorrect. Hence, $$f\left( x \right)$$ is not differentiable at $$x=\dfrac{\pi }{2},\pi $$ and $$\dfrac{3\pi }{2}$$. And differentiable at all other values in $$\left( 0,2\pi \right)$$ Answer is (c) Note: The left-hand derivative of $$f\left( x \right)$$at $$x=a$$is given as $${{L}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a-h \right)-f\left( a \right)}{-h}$$ and the right-hand derivative of $$f\left( x \right)$$at $$x=a$$is given as $${{R}^{'}}=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$$ . Students generally get confused between the two functions. Try to practice many questions based on these formulae so that confusion can be avoided.