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 \[\left| \sin x \right|=\left\{ \begin{align}
  & \sin x,0\le x\le \pi \\
 & -\sin x,\pi \le x\le 2\pi \\
\end{align} \right.\]
and \[\left| \cos x \right|=\left\{ \begin{align}
  & \cos x,\text{ }0\le x\le \dfrac{\pi }{2} \\
 & -\cos x,\text{ }\dfrac{\pi }{2}\le x\le \dfrac{3\pi }{2} \\
 & \cos x,\text{ }\dfrac{3\pi }{2}\le x\le 2\pi \\
\end{align} \right.\]
So, we can rewrite the function as,
\[f\left( x \right)=\left\{ \begin{align}
  & \sin x+\cos x,\text{ }0\le x\le \dfrac{\pi }{2} \\
 & \sin x-\cos x,\text{ }\dfrac{\pi }{2}\le x\le \pi \\
 & -\sin x-\cos x,\text{ }\pi \le x\le \dfrac{3\pi }{2} \\
 & -\sin x+\cos x,\text{ }\dfrac{3\pi }{2}\le x\le 2\pi \\
\end{align} \right.\]
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.