Question:
Which of the following functions is differentiable at $x = 0$ ?
Which of the following functions is differentiable at $x = 0$ ?
Updated On: Jun 21, 2022
- $\cos(|x|) + |x|$
- $\cos(|x|) - |x|$
- $\sin(|x|) + |x|$
- $\sin(|x|) - |x|$
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The Correct Option is D
Solution and Explanation
Let $f(x) = \sin(|x|) - |x|$
$ \Rightarrow f\left(x\right) = \begin{cases}
\sin \: x - x, x > 0 \\
- \sin \: x + x, x > 0
\end{cases}$
$ \Rightarrow f'\left(x\right) = \begin{cases}
\cos\: 1 - x, x \geq 0 \\
- \cos\: x + 1, x > 0
\end{cases}$
$ \therefore$ L.H.D = $\lim_{x\to0^{-}}\left(-\cos x+1\right) = -1+1=0$
R.H.D = $\lim_{x\to0^{+}}\left(\cos x - 1\right) = -1 - 1=0$
$ \because$ L.H.D = R.H.D
$ \therefore \:\:\: f(x)$ is differentiable at $x = 0$
$ \Rightarrow f\left(x\right) = \begin{cases}
\sin \: x - x, x > 0 \\
- \sin \: x + x, x > 0
\end{cases}$
$ \Rightarrow f'\left(x\right) = \begin{cases}
\cos\: 1 - x, x \geq 0 \\
- \cos\: x + 1, x > 0
\end{cases}$
$ \therefore$ L.H.D = $\lim_{x\to0^{-}}\left(-\cos x+1\right) = -1+1=0$
R.H.D = $\lim_{x\to0^{+}}\left(\cos x - 1\right) = -1 - 1=0$
$ \because$ L.H.D = R.H.D
$ \therefore \:\:\: f(x)$ is differentiable at $x = 0$
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.