If $f:R ightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{2\sin{x - \sin{2x}}}{2x\cos x} \ \ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ x = 0 \end{array} ight.\ ,\ if\ x
eq 0,$ Then the value of $a$ so that $f$ is continuous at 0 is

Question

If $f:R ightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{2\sin{x - \sin{2x}}}{2x\cos x} \ \ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ x = 0 \end{array} ight.\ ,\ if\ x 
eq 0,$ Then the value of $a$ so that $f$ is continuous at 0 is

Dr. Abubacker · Accepted Answer

(d)
$f(x) = \left\{ \begin{array}{r}
\frac{2\sin{x - \sin{2x}}}{2x\cos x} \
\ \ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ x = 0
\end{array} ight.\ ,\ if\ x 
eq 0,$
Now, $\lim_{x ightarrow 0}{f(x) = \lim_{x ightarrow 0}}\ \frac{2\sin{x - \sin{2x}}}{2x\cos x}$ $\left( \frac{0}{0}\ \ form ight)$
$= \lim_{x ightarrow 0}\ \ \frac{2\cos{x - 2\cos{2x}}}{2{(cos}x - x\sin{x)}}$
$= \ \lim_{x ightarrow 0}\ \ \frac{2 - 2}{2(1 - 0)} = 0$
Since, $f(x)$ is continuous at $x = 0$
$	herefore\ \ \ \ f(0) = \lim_{x ightarrow 0}\ f(x)$
$\Rightarrow a = 0$

Q. If $f:R \rightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} \frac{2\sin{x - \sin{2x}}}{2x\cos x} \\ \ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ x = 0 \end{array} \right.\ ,\ if\ x \neq 0,$ Then the value of $a$ so that $f$ is continuous at 0 is

Explanation

Reviewed by Dr. Abubacker

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