Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin 5x}{x^2+2x},& x\ne 0\\ k+\frac{1}{2},& x=0\end{array}$ is continuous at $x=0$, then the value of k is

• A. $1$
• B. $-2$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$2$

$f\left(x\right)=\{\begin{array}{cc}\frac{\sin 5x}{x^2+2x},& x\ne 0\\ k+\frac{1}{2},& x=0\end{array}$

LHL $f\left(0^{-}\right)=\underset{h\to 0}{lim}f(0-h)$

$=\underset{h\to 0}{lim}\frac{\sin 5(0-h)}{(0-h{)}^2+2(0-h)}$

$=\underset{h\to 0}{lim}\frac{\sin (-5h)}{h^2-2h}$

$=-\underset{h\to 0}{lim}\frac{\frac{\sin 5h}{5h}}{\frac{1}{5}(h-2)}=-\frac{1}{\frac{1}{5}(-2)}$

$=\frac{5}{2}$

Since, it is continuous at $x=0$

$\therefore LHL=f\left(0\right)$

$\Rightarrow \frac{5}{2}=k+\frac{1}{2}$

$\Rightarrow k=2$