Question

Let $f\left(x\right)=x+|x|+\cos \left(\right[{\pi }^2\left]x\right)$ and $g\left(x\right)=\sin x$, where $[.]$ denotes the greatest integer function. If $f\left(x\right)$ and $g\left(x\right)$ are defined in $[-\pi ,\pi ]$, then

• A. $f\left(x\right)/g\left(x\right)$ is continuous everywhere
• B. $f\left(x\right)$ is continuous everywhere
• C. $f\left(x\right)+g\left(x\right)$ is continuous everywhere
• D. $f\left(x\right)\cdot g\left(x\right)$ is continuous everywhere

Answer 1

Answer: (B), (C), (D)

$f\left(x\right)\cdot g\left(x\right)$ is continuous everywhere

$f\left(x\right)=x+|x|+\cos \left(\right[{\pi }^2\left]x\right)$, $g\left(x\right)=\sin x$
Since, both $f\left(x\right)$ and $g\left(x\right)$ are continuous everywhere, $f\left(x\right)+g\left(x\right)$ is continuous everywhere.

Now, $h\left(x\right)=f\left(x\right)\cdot g\left(x\right)$
$h\left(x\right)=(x+|x|+\cos 9x)\sin x$
Clearly, $h\left(x\right)$ is continuous
(as product of two continuous function is a continuous function)

Now, $k\left(x\right)=f\left(x\right)/g\left(x\right)$
$k\left(x\right)=\frac{x+|x|+\cos 9x}{\sin x}$
Clearly, $k\left(x\right)$ is not defined as $\sin x=0\text{at}x=0,\pm \pi$ .
So, it is discontinuous.