Question

If f(x) = $\{\begin{array}{ccc}-1,& if& x<0\\ 0,& if& x=0\\ 1,& if& x>0\end{array}$ and $g\left(x\right)=\sin x+\cos x$ then point of discontinuity of $\left(fog\right)\left(x\right)$ in $\left(0,2\pi \right)$ 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

The correct option is D $\frac{3\pi }{4},\frac{7\pi }{4}$

$f\left(x\right)=\{\begin{array}{c}-1,x<0\\ 0,x=0\\ 1,x>0\end{array}$

$g\left(x\right)=\sin x+\cos x$

$\left(fog\right)\left(x\right)=\{\begin{array}{c}-1,g\left(x\right)<0\\ 0,g\left(x\right)=0\\ 1,g\left(x\right)>0\end{array}$

We have $x\in \left(0,2\pi \right).$

We can rewrite $g\left(x\right)$ as:

$g\left(x\right)=\sin x+\cos x$

$=\sqrt{2}\left[\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right]$

$=\sqrt{2}\sin \left(\frac{\pi }{4}+x\right)$

Now, $g\left(x\right)=0$ when $\frac{\pi }{4}+x=\pi \Rightarrow x=\frac{3\pi }{4}$ and

$\frac{\pi }{4}+x=2\pi \Rightarrow x=\frac{7\pi }{4}$

Again, $g\left(x\right)<0$ when $\pi <\frac{\pi }{4}+x<2\pi$

Or, $\frac{3\pi }{4}<x<\frac{7\pi }{4}$

Also, $g\left(x\right)>0$ when $0<\frac{\pi }{4}+x<\pi$

Or, $-\frac{\pi }{4}<x<\frac{3\pi }{4}$

In the given domain, $g\left(x\right)>0$ when $0<x<\frac{3\pi }{4}$

Also, when $\frac{7\pi }{4}<x<2\pi$, we have

$2\pi <x+\frac{\pi }{4}<2n+\frac{\pi }{4}$

And $g\left(x\right)>0$

$\left(fog\right)\left(x\right)=\{\begin{array}{c}-1,\frac{3\pi }{4}<x<\frac{7\pi }{4}\\ 0,x=\frac{3\pi }{4},\frac{7\pi }{4}\\ 1,0<x<\frac{3\pi }{4}\&\frac{7\pi }{4}<x<2\pi \end{array}$

Thus, points of discontinuity: $\frac{3\pi }{4},\frac{7\pi }{4}$