Let (x) = {cases} (x-1)1/x-1, & {if } x ≠ 1 \\ 0, & {if } x = 1 {cases} Then which one of the following is true?

Correct Answer is : f is differentiable at x = 0 but not at x = 1

f is differentiable at x = 0 and x =1

f is differentiable at x = 0 but not at x = 1

f is differentiable at x = 1 but not at x = 0

f is neither differentiable at x = 0 nor at x =1

Correct Answer is : f is differentiable at x = 0 but not at x = 1

Test Index

BITSAT 2014 Paper

Section: Mathematics

Question : 108 of 150

Marks: +1, -0

Let (x) =

\begin{cases} (x-1)\sin\frac{1}{x-1}, & \text{if } x \neq 1 \\ 0, & \text{if } x = 1 \end{cases}

Then which one of the following is true?

Solution:

We have; f (x) =

\begin{cases} (x-1)\sin\frac{1}{x-1}, & \text{if } x \neq 1 \\ 0, & \text{if } x = 1 \end{cases}

R f ' (1) =

\lim\limits_{h\to 0}

\frac{f(1+h)-f(1)}{h}

\lim\limits_{h\to 0}

\frac{h\sin\frac{1}{h}-0}{h}

\lim\limits_{h\to 0}

sin

\frac{1}{h}

which does not exist.

∴ f is not differentiable at x = 1

Also

f ' (0) = sin

\frac{1}{x-1} - \frac{x-1}{(x-1)^2}

cos

\left.\frac{1}{x-1}\right|_{x=0}

= –sin 1 + cos 1

∴ f is differentiable at x = 0

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