Question

$f(x) = \begin{cases} \frac{sin(x-|x|)}{x-|x|} & \quad {x \in(-2,-1) } \\\ max{2x,3[|x|]}, & \quad \text{|x|<1}\\\1 & \quad \text{,otherwise} \end{cases}$ Where [t] denotes greatest integer t. If m is the number of points where f is not continuous and n is the number of points where f is not differentiable, then the ordered pair (m, n) is

• A. (3, 3)
• B. (2, 4)
• C. (2, 3)
• D. (3, 4)

Answer 1

Answer: (C)

(2, 3)

Answer 2

$f(x) = \begin{cases} \frac{sin(x-|x|)}{x-|x|} & \quad {x \in(-2,-1) } \\\ max{2x,3[|x|]}, & \quad \text{|x|<1}\\\1 & \quad \text{,otherwise} \end{cases}$

$\begin{cases} \frac{sin(x+2)}{x+2} & \quad {x \in(-2,-1) } \\\ 0, & \quad x \in(-1,0)]\\\1 & \quad \text{,otherwise} \end{cases}$

It clearly shows that $f(x)$ is discontinuous at $x = –1,$ $1$ also non differentiable and at $x = 0$,

L.H.D

= $\lim_{h\to0} \frac{f(0+h)-f(0)}{h}$ = $0$

R.H.D

$\lim_{h\to0} \frac{f(0+h)-f(0)}{h} =2$

∴ $f(x)$ is not differentiable at $x = 0$

∴ $m = 2$, $n = 3$

Hence, the correct option is (C): $(2, 3)$