Question

If $f\left(x\right)=\sin \left|x\right|$ and $g\left(x\right)=\tan \left|x\right|$ then discuss the continuity of $f\left(x\right)\pm g\left(x\right);\frac{f\left(x\right)}{g\left(x\right)}$ and $f\left(x\right).g\left(x\right)$

• A. $f\left(x\right)g\left(x\right)$ and $f\left(x\right)\pm g\left(x\right)$ are discontinuous at $x=(2n+1)\frac{\pi }{2};n\in 1$ $\frac{f\left(x\right)}{g\left(x\right)}$ is discontinuous at $x=\frac{n\pi }{2};n\in 1$
• B. $f\left(x\right)g\left(x\right)$ and $f\left(x\right)\pm g\left(x\right)$ are continuous at $x=(2n+1)\frac{\pi }{2};n\in 1$ $\frac{f\left(x\right)}{g\left(x\right)}$ is discontinuous at $x=\frac{n\pi }{2};n\in 1$
• C. $f\left(x\right)g\left(x\right)$ and $f\left(x\right)\pm g\left(x\right)$ are discontinuous at $x=(2n+1)\frac{\pi }{2};n\in 1$ $\frac{f\left(x\right)}{g\left(x\right)}$ is continuous at $x=\frac{n\pi }{2};n\in 1$
• D. $f\left(x\right)g\left(x\right)$ and $f\left(x\right)\pm g\left(x\right)$ are discontinuous at $x=(2n-1)\frac{\pi }{2};n\in 1$ $\frac{f\left(x\right)}{g\left(x\right)}$ is discontinuous at $x=\frac{\pi }{2};n\in 1$

Answer 1

Answer: (A)

$f\left(x\right)g\left(x\right)$ and $f\left(x\right)\pm g\left(x\right)$ are discontinuous at $x=(2n+1)\frac{\pi }{2};n\in 1$ $\frac{f\left(x\right)}{g\left(x\right)}$ is discontinuous at $x=\frac{n\pi }{2};n\in 1$

$f\left(x\right)=\sin \left|x\right|$
which is a composition of functions
Let $u\left(x\right)=|x|;v\left(x\right)=\sin x$
$\Rightarrow \left(vou\right)\left(x\right)=\sin |x|=f\left(x\right)$
Here,$R\left(u\right)\subseteq D\left(v\right)$ ($R\left(u\right)=R;D\left(v\right)=R$)
$vou$ is continuous function.
Hence, f(x) is continuous function.
Now,$g\left(x\right)=\tan \left|x\right|$
which is a composition of functions
Let $u\left(x\right)=|x|;v\left(x\right)=\tan x$
$\Rightarrow \left(vou\right)\left(x\right)=\tan |x|=g\left(x\right)$
Here,$R\left(u\right)⊈D\left(v\right)$ ($R\left(u\right)=R;D\left(v\right)=R-\left\{\right(2n+1\left)\frac{\pi }{2}\right\}$)
$vou$ is not continuous function.
Hence, g(x) is not continuous function.
So, the sum , difference and product all will be discontinuous at odd multiples of $\frac{\pi }{2}$