Question

lf $f\left(x\right)=\{\begin{array}{c}-1,x<0\\ 0,x=0\\ 1,x>0\end{array}$ and $g\left(x\right)=\sin x+\cos x$, then points of discontinuity of $\left(fog\right)\left(x\right)$ in $(0,2\pi )$ are

• A. $\frac{\pi }{4},\frac{5\pi }{4}$
• B. $\frac{\pi }{4},\frac{3\pi }{4}$
• C. $\frac{\pi }{4},\frac{7\pi }{4}$
• D. $\frac{3\pi }{4},\frac{7\pi }{4}$

Answer 1

Answer: (D)

$\frac{3\pi }{4},\frac{7\pi }{4}$

From the definition of $f\left(x\right)$, it is discontinuous at $x=0$.

Therefore, $fog\left(x\right)$ will be discontinuous whenever $g\left(x\right)=0$ in the interval $0<x<2\pi$.

The function $g\left(x\right)$ can be expressed as below:
$g\left(x\right)=\sin x+\cos x=\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)=\sqrt{2}(\cos \frac{\pi }{4}\sin x+\sin \frac{\pi }{4}\cos x)=\sqrt{2}\sin (x+\frac{\pi }{4})$

Now we need to determine the points where $g\left(x\right)=0$, since the function is discontinuous at those points:

$\sqrt{2}\sin (x+\frac{\pi }{4})=0\Rightarrow x+\frac{\pi }{4}=n\pi ,nϵZ\Rightarrow x=n\pi -\frac{\pi }{4}=\dots ,-\frac{\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4},\dots$

But $xϵ(0,2\pi )$. So the points of discontinuity in this interval are $x=\frac{3\pi }{4},\frac{7\pi }{4}$.