Question

The value of m for which the function $f(x) = \left\{ \begin{matrix} mx^{2},x \leq 1 \\ 2x,x > 1 \end{matrix} \right.\$ is continuous at $x = 1$, is

• A. 0
• B. 1
• C. 2
• D. Does not exist

Answer 1

Answer: (C)

2

Answer 2

LHL = $\lim_{x \rightarrow 1^{-}}f(x) = \lim_{h \rightarrow 0}m(1 - h)^{2} = m$

RHL = $\lim_{x \rightarrow 1^{+}}f(x) = \lim_{h \rightarrow 0}2(1 + h) = 2$ and $f(1) = m$

Function is continuous at $x = 1$, $x ^ { 2 } - 1 > 0$ LHL = RHL = $f(1)$

Therefore $m = 2$.