Question

$ If\, f(x) = \begin{cases} \frac {Sin \ 5X} {x^2+2x},X \neq 0 & \quad \\ k+ \frac {1} {2} ,x =0 & \quad \\ \end{cases} $
is continuous at $x = 0$ then the value of $k$ is

• A. 1
• B. -2
• C. 2
• D. 1/2

Answer 1

Answer: (C)

$f(x) = \begin{cases} \frac {Sin \ 5X} {x^2+2x},X \neq 0 & \quad \\ k+ \frac {1} {2} ,x =0 & \quad \\ \end{cases} $
L.H.L. $f\left(0^{-}\right)=\displaystyle\lim _{h \rightarrow 0} f(0-h)$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\sin 5(0-h)}{(0-h)^{2}+2(0-h)}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{\sin (-5 h)}{h^{2}-2 h}$
$=-\displaystyle\lim _{h \rightarrow 0} \frac{\frac{\sin 5 h}{5 h}}{\frac{1}{5}(h-2)}$
$=-\frac{1}{\frac{1}{5}(-2)}=\frac{5}{2}$
Since, it is continuous at $x=0$
$\therefore $ L.H.L. =f(0)
$\Rightarrow \,\frac{5}{2}=k+\frac{1}{2}$
$ \Rightarrow \,k=2$