If f(x)={cases} x e^{-≤ft(1/|x|+1/z)} & x ≠ 0 \\ 0 & x=0 {cases}. then f(x) is [AIEEE 2003]

Correct Answer is : continuous for all xxx but not differentiable at x=0x=0x=0

continuous as well as differentiable for all x

continuous for all x but not differentiable at x=0

neither differentiable nor continuous at x=0

Correct Answer is : continuous for all xxx but not differentiable at x=0x=0x=0

Test Index

Limits, Continuity and Differentiability Part 1

Section: Mathematics

Question : 89 of 100

Marks: +1, -0

f(x)=\begin{cases} x e^{-\left(\frac{1}{|x|}+\frac{1}{z}\right)} & x \neq 0 \\ 0 & x=0 \end{cases}.

f(x)

Solution:

f(0)=0 ; f(x)=x e^{-\left(\frac{1}{12\mid}+\frac{1}{x}\right)}

R. H.L.

\lim\limits_{h \rightarrow 0}(0+h) e^{-\frac{2}{h}}

=\lim\limits_{h \rightarrow 0}\frac{h}{e^{\frac{2}{h}}}=0

L.H.L.

\lim\limits_{h \rightarrow 0}(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0

therefore,

f(x)

is continuous,

R. H. D

=\lim\limits_{h \rightarrow 0}\frac{(0+h) e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-0}{h}=0

L.H.D.

=\lim\limits_{h \rightarrow 0}\frac{(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-0}{-h}=1

therefore, L.H.D.

\neq

R.H. D.

f(x)

is not differentiable at

x=0

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