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# Discuss the continuity of the following functions $(c)\;f(x)=\sin x.\cos x$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

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A)
Toolbox:
• $\sin 2A=2\sin A\cos A$
Step 1:
Here $f(x)=\sin x\cos x$
Multiply and divide by 2
$f(x)=\large\frac{1}{2}$$\times 2\sin x\cos x \qquad=\large\frac{1}{2}$$\sin 2x$
We know that $\sin 2A=2\sin A\cos A$
Step 2:
At $x=a$ where $a\in R$
LHL =$\lim\limits_{x\to a^-}f(x)=\lim\limits_{x\to a^-}\large\frac{1}{2}$$\sin 2x \qquad=\lim\limits_{h\to 0}\large\frac{1}{2}$$\sin 2(a-h)$
$\qquad=\large\frac{1}{2}$$[\sin 2a\cos 2h-\cos 2a\sin 2h] \qquad=\large\frac{1}{2}$$[\sin 2a\cos 0-\cos 2a\sin 0]$
$\qquad=\large\frac{1}{2}$$\sin 2a Step 3: LHL =\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^-}\large\frac{1}{2}$$\sin 2x$
$\qquad=\lim\limits_{h\to 0}\large\frac{1}{2}$$\sin 2(a+h) \qquad=\large\frac{1}{2}$$[\sin 2a\cos 2h+\cos 2a\sin 2h]$
$\qquad=\large\frac{1}{2}$$[\sin 2a\cos 0+\cos 2a\sin 0] \qquad=\large\frac{1}{2}$$\sin 2a$
Step 4:
$f(a)=\large\frac{1}{2}$$\sin 2a$
$\therefore$ LHL =RHL=f(a).
Hence $f(x)$ is continuous at all points.