Intersection of graphs, and no solution for trig functions.

Question

All I know is the $c=a\sin(x-b) $.

I don't know how to check the values of $b$ for "no solutions", in the case of trig. functions.

Can someone people provide an algebraic method to solve this?

Answer 1

First of all, if $f(x)=a\sin(x-b)$, then $-a\le f(x)\le a$ for all $a,b,x\in\mathbb{R}$, $a>0$.

Then, in particular, $-a\le f(x)\le a$ for all $a,b>0$ and $0\le x\le\pi$.

Then it is required that $f(x)=c>0$, for some $x$ such that $0\le x\le\pi$.

Now,

1. $x\in[\,0,\pi\,]\ ,\ b=0\ \Rightarrow\ x-b\in[\,0,\pi\,]$;
2. $x\in[\,0,\pi\,]\ ,\ b=\frac{\pi}{4}\ \Rightarrow\ x-b\in[\,-\frac{\pi}{4},\frac{3\pi}{4}\,]$;
3. $x\in[\,0,\pi\,]\ ,\ b=\frac{\pi}{2}\ \Rightarrow\ x-b\in[\,-\frac{\pi}{2},\frac{\pi}{2}\,]$;
4. $x\in[\,0,\pi\,]\ ,\ b=\pi\ \Rightarrow\ x-b\in[\,-\pi,0\,]$;
5. $x\in[\,0,\pi\,]\ ,\ b=2\pi\ \Rightarrow\ x-b\in[\,-2\pi,-\pi\,]=[\,0,\pi\,]$.

Case 4. (D in your question) is the only one where $f(x)$ is always negative: $a\sin\left([\,-\pi,0\,]\right)=[\,-a,0\,]$ and $f(x)=c>0$ has no solution here, hence the answer is D.