Question

If $f : R \to R$ is defined by
$f(x) = \begin{cases} \frac{ 2 \ \sin \ x - \sin \ 2x}{ 2x \ \cos \ x} & \text{if } x \neq 0 \\ a , & \text{if } x = 0 \end{cases}$ 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, $f(x)= \begin{cases}\frac{2 \sin (x)-\sin (2 x)}{2 x \cos (x)}, & \text { if } x \neq 0 \\ a, \text { if } x=0, & \text { if } x=0\end{cases}$
Now, $\displaystyle\lim _{x \rightarrow 0} f(x)=\displaystyle\lim _{x \rightarrow 0} \frac{2 \sin (x)-\sin (2 x)}{2 x \cos (x)} \left[\frac{0}{0} \text { form }\right]$
$= \displaystyle\lim _{x \rightarrow 0} \frac{2 \cos (x)-2 \cos (2 x)}{2(\cos (x)-x \sin (x))} $
$=\displaystyle\lim _{x \rightarrow 0} \frac{2-2}{(1-0)}=0$
Since, $f(x)$ is continuous at $x=0$
$\therefore f(0)=\displaystyle\lim _{x \rightarrow 0} f(x) \Rightarrow a=0$