Question

If $f:R\to R$ is defined by
$f(x)=\{\begin{array}{ll}\frac{2\sin x-\sin 2x}{2x\cos x}& \text{if }x\ne 0\\ a,& \text{if }x=0\end{array}$ then the value of $a$ so that $f$ is continuous at $0$ is

• A. 2
• B. 1
• C. -1
• D. 0

Answer 1

Answer: (D)

Given, $f(x)=\{\begin{array}{ll}\frac{2\sin (x)-\sin (2x)}{2x\cos (x)},& \text{ if }x\ne 0\\ a,\text{ if }x=0,& \text{ if }x=0\end{array}$
Now, $\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{2\sin (x)-\sin (2x)}{2x\cos (x)}\left[\frac{0}{0}\text{ form }\right]$
$=\underset{x\to 0}{\lim }\frac{2\cos (x)-2\cos (2x)}{2(\cos (x)-x\sin (x))}$
$=\underset{x\to 0}{\lim }\frac{2-2}{(1-0)}=0$
Since, $f(x)$ is continuous at $x=0$
$\therefore f(0)=\underset{x\to 0}{\lim }f(x)\Rightarrow a=0$