Question

if $f(n) \begin{cases} \frac{1-\cos \,K x}{x \,\sin x}, & \text{if $x\ne0$} \\ \frac{1}{2}, & \text{if x=0} \end{cases}$
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 \displaystyle\lim _{x \rightarrow 0} f(x)=f(0) $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{1-\cos\, K x}{x\, \sin\, x}=\frac{1}{2}$
Applying L' Hopitals' rule, we get
$\displaystyle\lim _{x \rightarrow 0} \frac{K\, \sin\, K x}{\sin\, x+x\, \cos\, x}=\frac{1}{2}$
Again, by L' Hopitals' rule,
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{K^{2} \cos \,K x}{\cos \,x-x\, \sin\, x+\cos \,x}\right)=\frac{1}{2} $
$\Rightarrow \frac{K^{2}}{1-0+1}=\frac{1}{2}$
$\Rightarrow K=\pm 1$