Question

Let $f\left(x\right)=\{\begin{array}{c}3x-4,0\le x\le 2\\ 2x+\lambda ,2<x\le 3\end{array}$. If $f$ is continuous at $x=2$, then $\lambda$ is-

• A. $-1$
• B. $0$
• C. $-2$
• D. $2$

Answer 1

The correct option is C $-2$

$f\left(x\right)=\{\begin{array}{c}3x-4,0\le x\le 2\\ 2x+\lambda ,2<x\le 3\end{array}$
LHL $=\underset{x\to 2^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(2-h)$
$=\underset{h\to 0}{lim}3(2-h)-4$
$=2$
RHL $=\underset{x\to 2^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(2+h)$
$=\underset{h\to 0}{lim}2(2+h)+\lambda$
$=4+\lambda$
Since, $f\left(x\right)$ is continuous at $x=2$,
$LHL=RHL=f\left(2\right)$
$\Rightarrow 4+\lambda =2$
$\Rightarrow \lambda =-2$