Question

Let $f\left(x\right)=\left\{\begin{array}{r}\sin \left[\frac{|x^2-5x+6|}{x^2-5x+6}\right]x\ne 2,3\\ 1x=2\mathrm{or}3\end{array}\right\}$ , the set of all points where $f$ is differentiable is

• A. $\left(-\infty ,\infty \right)$
• B. $\left(-\infty ,\infty \right)-\left\{2\right\}$
• C. $\left(-\infty ,\infty \right)-\left\{3\right\}$
• D. $\left(-\infty ,\infty \right)-\left\{2,3\right\}$

Answer 1

The correct option is D

$\left(-\infty ,\infty \right)-\left\{2,3\right\}$

Explanation for Correct Answer:

Use limits to find the differentiable points:

Given that:

$f\left(x\right)=\left\{\begin{array}{r}\sin \left[\frac{|x^2-5x+6|}{x^2-5x+6}\right]x\ne 2,3\\ 1x=2\mathrm{or}3\end{array}\right\}$

The sine function is always differentiable. Also, $1$ is constant and hence always differentiable.

Hence the function may or may not be differentiable at only two points that are $2,3$.

Let's check at $x=2$.

$\begin{array}{l}{f}^{\text{'}}\left(2^{+}\right)=\underset{x\to 0}{\lim }\frac{f(2+h)+f\left(2\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{h(h-1)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin (h-1)}{h^2(h-1)}\\ =-\infty \end{array}$

Let's check at $x=3$.

$\begin{array}{l}{f}^{\text{'}}\left(3^{+}\right)=\underset{x\to 0}{\lim }\frac{f(3+h)+f\left(3\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{h(h-1)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin (h-1)}{h^2(h-1)}\\ =-\infty \end{array}$

We can say that $x$ is not differentiable at $x=2\text{and}3$.

Therefore, The required set is $\left(-\infty ,\infty \right)-\left\{2,3\right\}$

Therefore, the correct answer is option (D).