The function f(x)=x^{-1}1/x for x≠ 0, f(0)=0 is

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

neither continuous at x=0 nor differentiable at x=0

continuous at x=0 but not differentiable at x=0

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

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UPSEE 2017 Solved Paper

Question : 143 of 150

Marks: +1, -0

The function

f(x)=x\tan^{-1}\frac{1}{x}

x\neq 0

f(0)=0

Solution:

Given,

f(x)=x\tan^{-1}\frac{1}{x}

for

x\neq 0

and

f(0)=0

\because -\frac{\pi}{2}\leq \tan^{-1}\frac{1}{x}\leq \frac{\pi}{2}

\therefore -\frac{\pi}{2}x\leq x\tan^{-1}\frac{1}{x}\leq \frac{\pi}{2}x

Here,

\lim\limits_{x\to 0}x\tan^{-1}\frac{1}{x}

=\lim\limits_{x\to 0}\frac{\tan^{-1}\frac{1}{x}}{\frac{1}{x}}

=0

\left[\because \lim\limits_{x\to 0}\frac{x}{t}\right]

And

f(0)=0 (

given

)

\therefore f(x)

is continuous at

x=0

But,

\lim\limits_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\to 0}\tan^{-1}\frac{1}{x}

does not exist.

\therefore f(x)

is not differential at

x=0

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