Question

if $f(n)\{\begin{array}{ll}\frac{1-\cos Kx}{x\sin x},& \text{if x=0}\\ \frac{1}{2},& \text{if x=0}\end{array}$
is continuous at x=0, then the value of K is

• A. $\pm \frac{1}{2}$
• B. 0
• C. $\pm 2$
• D. $\pm 1$

Answer 1

Answer: (D)

Given, $f$ is continuous at $x=0$.
$\Rightarrow \underset{x\to 0}{\lim }f(x)=f(0)$
$\Rightarrow \underset{x\to 0}{\lim }\frac{1-\cos Kx}{x\sin x}=\frac{1}{2}$
Applying L' Hopitals' rule, we get
$\underset{x\to 0}{\lim }\frac{K\sin Kx}{\sin x+x\cos x}=\frac{1}{2}$
Again, by L' Hopitals' rule,
$\underset{x\to 0}{\lim }\left(\frac{K^2\cos Kx}{\cos x-x\sin x+\cos x}\right)=\frac{1}{2}$
$\Rightarrow \frac{K^2}{1-0+1}=\frac{1}{2}$
$\Rightarrow K=\pm 1$