Question

$Iff(x)=\{\begin{array}{ll}\frac{Sin5X}{x^2+2x},X\ne 0& \\ k+\frac{1}{2},x=0& \end{array}$
is continuous at $x=0$ then the value of $k$ is

• A. 1
• B. -2
• C. 2
• D. 1/2

Answer 1

Answer: (C)

$f(x)=\{\begin{array}{ll}\frac{Sin5X}{x^2+2x},X\ne 0& \\ k+\frac{1}{2},x=0& \end{array}$
L.H.L. $f\left(0^{-}\right)=\underset{h\to 0}{\lim }f(0-h)$
$=\underset{h\to 0}{\lim }\frac{\sin 5(0-h)}{(0-h{)}^2+2(0-h)}$
$=\underset{h\to 0}{\lim }\frac{\sin (-5h)}{h^2-2h}$
$=-\underset{h\to 0}{\lim }\frac{\frac{\sin 5h}{5h}}{\frac{1}{5}(h-2)}$
$=-\frac{1}{\frac{1}{5}(-2)}=\frac{5}{2}$
Since, it is continuous at $x=0$
$\therefore$ L.H.L. =f(0)
$\Rightarrow \frac{5}{2}=k+\frac{1}{2}$
$\Rightarrow k=2$