Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

The function $f(x)=\begin{Bmatrix}\frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}};&x≠0\\0;&x=0\end{Bmatrix}$ is:

Options:

Continuous everywhere but not differentiable

Continuous and differentiable everywhere

Not continuous at x = 0

None of these

Correct Answer:

Continuous everywhere but not differentiable

Explanation:

$f(x)=\begin{Bmatrix}\frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}};&x≠0\\0;&x=0\end{Bmatrix}$

$LHD|_{x=0}=\underset{h→0}{\lim}\frac{(-h)\frac{(e^{-1/h}-e^{1/h})}{e^{-1/h}+e^{1/h}}}{-h}=\underset{h→0}{\lim}\frac{e^{-2/h}-1}{e^{-2/h}+1}=-1$

$RHD|_{x=0}=\underset{h→0}{\lim}\frac{(h)\frac{e^{1/h}-e^{-1/h}}{e^{-1/h}+e^{-1/h}}-0}{h}=\underset{h→0}{\lim}\frac{1-e^{-2/h}}{1+e^{-2/h}}=1$

Function is non-diff. but continuous as LHD and RHD exists