Question

$f\left(x\right)=\{\begin{array}{cc}\left[x\right]+[-x],& \\ \lambda ,& \end{array}\begin{array}{cc}x\ne 2& \\ x=2& \end{array},$ then f(x) is continuous at $x=2$ provided $\lambda$ is:

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $-1$

$RHL={lim}_{x\to 2^{+}}\left[x\right]+[-x]$

$={lim}_{h\to 0}[2+h]+[-(2+h\left)\right]$

$=2-3$

$\Rightarrow RHL=-1$

Now, $LHL={lim}_{x\to 2^{-}}\left[x\right]+[-x]$

$={lim}_{h\to 0}[2-h]+[-(2-h\left)\right]$

$=1-2$

$\Rightarrow LHL=-1$

Since, $f\left(x\right)$ is continuous at $x=2$

$LHL=RHL=f\left(2\right)$

$\Rightarrow \lambda =-1$