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If $f: R \rightarrow R$ is defined by
$f(x)=\left\{\begin{array}{cc}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text { if } x \neq 0 \\ a, & \text { if } x=0\end{array}\right.$
then the value of $a$ so that $f$ is continuous at 0 is
Options:
$f(x)=\left\{\begin{array}{cc}\frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text { if } x \neq 0 \\ a, & \text { if } x=0\end{array}\right.$
then the value of $a$ so that $f$ is continuous at 0 is
Solution:
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Verified Answer
The correct answer is:
0
$$
\begin{array}{l}
\text { (d) } f(x)=\left\{\begin{array}{cc}
\frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text { if } x \neq 0 \\
a, & \text { if } x=0
\end{array}\right. \\
\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x}\left(\frac{0}{0} \text { form }\right) \\
=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}
\end{array}
$$
$$
=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0
$$
Since, $f(x)$ is contineous at $x=0$
$\therefore f(0)=\lim _{x \rightarrow 0} f(x) \Rightarrow a=0$
\begin{array}{l}
\text { (d) } f(x)=\left\{\begin{array}{cc}
\frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text { if } x \neq 0 \\
a, & \text { if } x=0
\end{array}\right. \\
\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{2 x \cos x}\left(\frac{0}{0} \text { form }\right) \\
=\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{2(\cos x-x \sin x)}
\end{array}
$$
$$
=\lim _{x \rightarrow 0} \frac{2-2}{2(1-0)}=0
$$
Since, $f(x)$ is contineous at $x=0$
$\therefore f(0)=\lim _{x \rightarrow 0} f(x) \Rightarrow a=0$
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