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

x2-5x+6 0⇒x 3 f(x)=sin(x2-5x+6)x2-5x+6 if x 3 Here L.H.L=limx→2sin(x2-5x+6)x2-5x+6=1 R.H.L=limx→2+sin(-(x2-5x+6))x2-5x+6=-1 ∴L.H.L≠R.H.L similarly at x=3 ,L.H.L≠R.H.L ∴ f(x) is not continuous at x=2,3The function is clearly differentiable except possible at x = 2,3

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

x2-5x+6 0⇒x 3 f(x)=sin(x2-5x+6)x2-5x+6 if x 3 Here L.H.L=limx→2sin(x2-5x+6)x2-5x+6=1 R.H.L=limx→2+sin(-(x2-5x+6))x2-5x+6=-1 ∴L.H.L≠R.H.L similarly at x=3 ,L.H.L≠R.H.L ∴ f(x) is not continuous at x=2,3The function is clearly differentiable except possible at x = 2,3

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Let $f (x) = \{\begin{matrix} \frac{\sin |x^{2} - 5 x + 6|}{x^{2} - 5 x + 6}, x \neq 2, 3 \\ 1, x = 2 or 3 \end{matrix}$ the set of all points where f is differentiable is

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$(- \infty, \infty)$

$(- \infty, \infty) ~ {2}$

$(- \infty, \infty) - {3}$

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

answer is D.

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Detailed Solution

$x^{2} - 5 x + 6 < 0 \Rightarrow 2 < x < 3 x^{2} - 5 x + 6 > 0 \Rightarrow x < 2 o r x > 3 f (x) = \{\begin{cases} \frac{\sin (x^{2} - 5 x + 6)}{x^{2} - 5 x + 6} i f x < 2 \\ 1 i f x = 2 \\ \frac{\sin (- (x^{2} - 5 x + 6))}{x^{2} - 5 x + 6} i f 2 < x < 3 \\ 1 i f x = 3 \\ \frac{\sin (x^{2} - 5 x + 6)}{x^{2} - 5 x + 6} i f x > 3 \end{cases} H e r e L . H . L = \lim_{x \to 2} \frac{\sin (x^{2} - 5 x + 6)}{x^{2} - 5 x + 6} = 1 R . H . L = \lim_{x \to 2^{+}} \frac{\sin (- (x^{2} - 5 x + 6))}{x^{2} - 5 x + 6} = - 1 ∴ L . H . L \neq R . H . L s i m i l a r l y a t x = 3, L . H . L \neq R . H . L ∴ f (x) i s n o t c o n t i n u o u s a t x = 2, 3$