Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin x}{x}& x\le 0\\ k-1& x>0\end{array}$ is continuous at $x=0$ then the value of $k$ is-

• A. 2
• B. 0
• C. -1
• D. 1

Answer 1

The correct option is A 2

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin x}{x}& x\le 0\\ k-1& x>0\end{array}$

Therefore,

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{\sin x}{x}=1$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}(k-1)=k-1$

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\sin x}{x}=1$

Given that $f\left(x\right)$ is continuous at $x=0$.

Thus,

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0}f\left(x\right)$

$\therefore 1=k-1$

$\Rightarrow k=1+1=2$

Hence the value of $k$ is $2$.