Let f(x)=sin⁡x2−5x+6×2−5x+6x≠2,31x=2 or 3the set of all points where f is differentiable is

# Let the set of all points where is differentiable is

1. A

$\left(-\mathrm{\infty },\mathrm{\infty }\right)$

2. B

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

3. C

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

4. D

$\left(-\mathrm{\infty },\mathrm{\infty }\right)~\left\{2,3\right\}$ or $R-\left\{2,3\right\}$

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### Solution:

The function is clearly differentiable except

possibly at $x=2,3$

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(2+\right)=\underset{h\to 0+}{lim} \frac{f\left(2+h\right)-f\left(2\right)}{h}\\ =\underset{h\to 0+}{lim} \frac{\mathrm{sin}h\left(1-h\right)+h\left(1-h\right)}{{h}^{2}\left(-1+h\right)}\\ =-\underset{h\to 0+}{lim} \left(\frac{\mathrm{sin}h\left(1-h\right)}{h\left(1-h\right)}+1\right)\frac{1}{h}\end{array}$

The last limit doesn’t exist. If this limit then $\underset{h\to 0+}{lim} \frac{1}{h}$ exist,

which is not true

Thus f is not differentiable at $x=2$ . Similarly f is not dif-

ferentiable   at . Hence the set of all points where f is

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