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Which of the following function(s) is differentiable at x = 0?(a) $\cos \left( \left| x \right| \right)+\left| x \right|$(b) $\cos \left( \left| x \right| \right)-\left| x \right|$(c) $\sin \left( \left| x \right| \right)+\left| x \right|$(d) $\sin \left( \left| x \right| \right)-\left| x \right|$

Last updated date: 15th Jul 2024
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Hint: Differentiate the given function and check $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$= finite quantity for differentiability at x = 0.

Here we have to check if the function given is differentiable at x = 0 or not.
We know that for the function to be differentiable at x = 0, it must satisfy $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$= finite quantity$....\left( i \right)$
We know that $\left| x \right|=x\text{ for }x\ge 0....\left( ii \right)$
And, $\left| x \right|=-x\text{ for }x<0....\left( iii \right)$
Taking $f\left( x \right)=\cos \left( \left| x \right| \right)+\left| x \right|$
By equation (ii) and (iii)
f\left( x \right)=\left\{ \begin{align} & \cos \left( x \right)+x,\text{ }x\ge 0 \\ & \cos \left( -x \right)-x,\text{ }x<0 \\ \end{align} \right.
As we know that $\cos \left( -x \right)=\cos x$
We get, f\left( x \right)=\left\{ \begin{align} & \cos x+x,\text{ }x\ge 0 \\ & \cos x-x,\text{ }x<0 \\ \end{align} \right.
Now, we will differentiate $f\left( x \right)$
Since, we know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$and $\dfrac{d}{dx}\left( x \right)=1$
We get, {{f}^{'}}\left( x \right)=\left\{ \begin{align} & -\sin x+1\text{, }x>0 \\ & -\sin x-1,\text{ x0} \\ \end{align} \right.
Now, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\sin x-1 \right)=-\sin \left( {{0}^{o}} \right)-1$ [Since $\sin \left( {{0}^{o}} \right)=0$]
We get $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=-1$
Now, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( -\sin x+1 \right)=-\sin \left( {{0}^{o}} \right)+1=1$
Since, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)$
Therefore, $f\left( x \right)=\cos \left( \left| x \right| \right)+\left| x \right|$is not differentiable at x = 0.
Now taking $g\left( x \right)=\cos \left( \left| x \right| \right)-\left| x \right|$
By equation (ii) and (iii)
We get, g\left( x \right)=\left\{ \begin{align} & cos x-x,\text{ }x\ge 0 \\ & \cos x+x,\text{ }x<0 \\ \end{align} \right.
Now, we will differentiate $g\left( x \right)$.
We get, {{g}^{'}}\left( x \right)=\left\{ \begin{align} & -\sin x-1,\text{ }x>0 \\ & -\sin x+1,\text{ }x<0 \\ \end{align} \right.\text{ }\!\![\!\!\text{ Also }\sin \left( {{0}^{o}} \right)\text{=0 }\!\!]\!\!\text{ }
Now, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\sin x+1 \right)=-\sin \left( {{0}^{o}} \right)+1=1$
And $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( -\sin x-1 \right)=-\sin \left( {{0}^{o}} \right)-1=-1$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{g}^{'}}\left( x \right)$
Therefore, $g\left( x \right)=\cos \left( \left| x \right| \right)-\left| x \right|$is not differentiable at x = 0.
Now, taking $h\left( x \right)=\sin \left( \left| x \right| \right)+\left| x \right|$
By equation (ii) and (iii),
We get h\left( x \right)=\left\{ \begin{align} & \sin x+x,\text{ }x\ge 0 \\ & \sin \left( -x \right)-x,\text{ }x<0 \\ \end{align} \right.\text{ }\!\![\!\!\text{ Since }\sin \left( -x \right)\text{=}-\sin x\text{ }\!\!]\!\!\text{ }
h\left( x \right)=\left\{ \begin{align} & \sin x+x,\text{ }x\ge 0 \\ & -\sin x-x,\text{ }x<0 \\ \end{align} \right.
Now, we will differentiate $h\left( x \right)$.
Since, $\dfrac{d}{dx}\left( \sin x \right)=\cos x$
We get, {{h}^{'}}\left( x \right)=\left\{ \begin{align} & \cos x+1,\text{ }x>0 \\ & -\cos x-1,\text{ }x<0 \\ \end{align} \right.\text{ }\!\![\!\!\text{ Also }\cos \left( {{0}^{o}} \right)\text{=1 }\!\!]\!\!\text{ }
Now, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\cos x-1 \right)=-\cos \left( {{0}^{o}} \right)-1=-1-1=-2$
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \cos x+1 \right)=\cos \left( {{0}^{o}} \right)+1=1+1=2$
Since, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{h}^{'}}\left( x \right)$
Therefore, $h\left( x \right)=\sin \left( \left| x \right| \right)+\left| x \right|$is not differentiable at x = 0.
Now, taking $J\left( x \right)=\sin \left( \left| x \right| \right)-\left| x \right|$
By equation (ii) and (iii),
We get J\left( x \right)=\left\{ \begin{align} & \sin x-x,\text{ }x\ge 0 \\ & \sin \left( -x \right)+x,\text{ }x<0 \\ \end{align} \right.\text{ }\!\![\!\!\text{ Since }\sin \left( -x \right)\text{=}-\sin x\text{ }\!\!]\!\!\text{ }
J\left( x \right)=\left\{ \begin{align} & \sin x-x,\text{ }x\ge 0 \\ & -\sin x+x,\text{ }x<0 \\ \end{align} \right.
Now, we will differentiate $J\left( x \right)$.
Since, $\dfrac{d}{dx}\left( \sin x \right)=\cos x$
We get, {{J}^{'}}\left( x \right)=\left\{ \begin{align} & \cos x-1,\text{ }x>0 \\ & -\cos x+1,\text{ }x<0 \\ \end{align} \right.\text{ }\!\![\!\!\text{ Also }\cos \left( {{0}^{o}} \right)\text{=1 }\!\!]\!\!\text{ }
Now, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \cos x-1 \right)=\cos \left( {{0}^{o}} \right)-1=1-1=0$
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -\cos x+1 \right)=-\cos \left( {{0}^{o}} \right)+1=-1+1=0$
Since, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{J}^{'}}\left( x \right)$
Therefore, J(x) is differentiable at x = 0.
Hence, option (d) is correct.

Note: Some students check the continuity first in these types of questions but that is not required. It makes the solution time consuming, because if a function is differentiable, it would surely be continuous as well and need not be checked.