Let f( x ) = \ \beginmatrix ( x - 1 )\sin 1 \over x - 1 & if\,x e 1 0 & if\,x = 1 \endmatrix . Then which one of the following is true?

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

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

f is differentiable at x = 0 and at x = 1

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

Limits Continuity and Differentiability PYQ — JEE Main 2008 · Shift 0 · Q31 with Solution

View written solutionFree

Correct answer: C

Given function

f(x)= \begin{cases} (x-1)\sin\left(\dfrac{1}{x-1}\right), & x\neq 1 \\ 0, & x=1 \end{cases}

We need to check differentiability at $x=0$ and $x=1$ .

Differentiability at $x=0$

Since $0\neq 1$ , in a neighborhood of $x=0$ the function is simply

f(x)=(x-1)\sin\left(\frac{1}{x-1}\right).

This is a product/composition of differentiable functions at $x=0$ because $x-1\neq 0$ there, so $\dfrac{1}{x-1}$ is well-defined and differentiable.

Hence $f$ is differentiable at $x=0$ .

For confirmation, we can even differentiate:

f'(x)=\sin\left(\frac{1}{x-1}\right)+(x-1)\cos\left(\frac{1}{x-1}\right)\left(-\frac{1}{(x-1)^2}\right)

f'(x)=\sin\left(\frac{1}{x-1}\right)-\frac{\cos\left(\frac{1}{x-1}\right)}{x-1}.

At $x=0$ , this exists and is finite. Therefore, $f$ is differentiable at $x=0$ .

Differentiability at $x=1$

First check continuity at $x=1$ :

\lim_{x\to 1}(x-1)\sin\left(\frac{1}{x-1}\right).

Using $|\sin t|\le 1$ ,

\left|(x-1)\sin\left(\frac{1}{x-1}\right)\right|\le |x-1|\to 0.

Thus,

\lim_{x\to 1} f(x)=0=f(1),

so $f$ is continuous at $x=1$ .

Now check derivative at $x=1$ :

f'(1)=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} =\lim_{x\to 1}\frac{(x-1)\sin\left(\frac{1}{x-1}\right)}{x-1} =\lim_{x\to 1}\sin\left(\frac{1}{x-1}\right).

But as $x\to 1$ , $\dfrac{1}{x-1}\to \pm\infty$ , and

\sin\left(\frac{1}{x-1}\right)

does not have a limit because it oscillates between $-1$ and $1$ .

Hence $f'(1)$ does not exist.

So $f$ is not differentiable at $x=1$ .

Conclusion

Differentiable at $x=0$ : Yes
Differentiable at $x=1$ : No

Therefore, the correct option is:

\boxed{\text{C: } f \text{ is differentiable at } x=0 \text{ but not at } x=1}

Comparison with stored answer

Stored correct answer: $\text{C}$

Our derived answer: $\text{C}$

They agree.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: C

Given function

f(x)= \begin{cases} (x-1)\sin\left(\dfrac{1}{x-1}\right), & x\neq 1 \\ 0, & x=1 \end{cases}

We need to check differentiability at $x=0$ and $x=1$ .

Differentiability at $x=0$

Since $0\neq 1$ , in a neighborhood of $x=0$ the function is simply

f(x)=(x-1)\sin\left(\frac{1}{x-1}\right).

This is a product/composition of differentiable functions at $x=0$ because $x-1\neq 0$ there, so $\dfrac{1}{x-1}$ is well-defined and differentiable.

Hence $f$ is differentiable at $x=0$ .

For confirmation, we can even differentiate:

f'(x)=\sin\left(\frac{1}{x-1}\right)+(x-1)\cos\left(\frac{1}{x-1}\right)\left(-\frac{1}{(x-1)^2}\right)

f'(x)=\sin\left(\frac{1}{x-1}\right)-\frac{\cos\left(\frac{1}{x-1}\right)}{x-1}.

At $x=0$ , this exists and is finite. Therefore, $f$ is differentiable at $x=0$ .

Differentiability at $x=1$

First check continuity at $x=1$ :

\lim_{x\to 1}(x-1)\sin\left(\frac{1}{x-1}\right).

Using $|\sin t|\le 1$ ,

\left|(x-1)\sin\left(\frac{1}{x-1}\right)\right|\le |x-1|\to 0.

Thus,

\lim_{x\to 1} f(x)=0=f(1),

so $f$ is continuous at $x=1$ .

Now check derivative at $x=1$ :

f'(1)=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} =\lim_{x\to 1}\frac{(x-1)\sin\left(\frac{1}{x-1}\right)}{x-1} =\lim_{x\to 1}\sin\left(\frac{1}{x-1}\right).

But as $x\to 1$ , $\dfrac{1}{x-1}\to \pm\infty$ , and

\sin\left(\frac{1}{x-1}\right)

does not have a limit because it oscillates between $-1$ and $1$ .

Hence $f'(1)$ does not exist.

So $f$ is not differentiable at $x=1$ .

Conclusion

Differentiable at $x=0$ : Yes
Differentiable at $x=1$ : No

Therefore, the correct option is:

\boxed{\text{C: } f \text{ is differentiable at } x=0 \text{ but not at } x=1}

Comparison with stored answer

Stored correct answer: $\text{C}$

Our derived answer: $\text{C}$

They agree.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

More from Limits Continuity and Differentiability