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

The Correct Option is C

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

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

 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): \((2, 3)\)