# Finding the range of values that a function $f(x,t)$ can take without tedious calculations

A few days ago, I found the following question in a book only with the answer:

Question : Letting $x, t$ be real numbers, then a function $f(x,t)$ is defined as $$f(x,t)=\frac{(2-2\cos x)t^2+4-2\cos x}{(1-2\sin x)t^2+2t+1-2\sin x}.$$

Then, find the range of values that $f(x,t)$ can take.

The answer is $f(x,t)\le -\frac34$ or $f(x,t)\gt0.$

This book says,"Tedious calculations are not needed.". I've been looking for a way without tedious calculations, but I'm facing difficulty. I need your help.

• The answer looks like a contradiction in itself. Please check. – Macavity Sep 15 '13 at 16:29
• @Macavity: Thank you for pointing it out. I edited it. – mathlove Sep 15 '13 at 16:52

I've just got the following:

Since $$f(x,t)=\frac{\frac{1-t^2}{1+t^2}-(-3+2\cos x)}{\frac{2t}{1+t^2}-(-1+2\sin x)},$$

we know that $f(x,t)$ is the slope of the line $PQ$ where $$P\left(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2}\right), Q\left(-1+2\sin x,-3+2\cos x\right).$$

We can see that $Q$ moves on a circle : $(x+1)^2+(y+3)^2=2^2$ and that $P$ moves on the unit circle whose center is the origin without a point $(0,-1)$.

By drawing these figures, we can get the result.

Here is a partial answer, a bit too long to include as comment. We can rewrite

$$f(x,t)=\frac{(1-\cos x)(t^2+1)+1}{(\frac{1}{2}-\sin x)(t^2+1)+t}$$

Now, as the numerator is clearly positive, $0$ is not in the range.

$f(0,t)=\dfrac{2}{(t+1)^2}$, so all positive values are in the range.

As $f(\frac{\pi}{2}, t) = -2\dfrac{t^2 + 2}{(t-1)^2}$ it is evident that all values $< -2$ are also in the range.

So we are left to search in $[-2, 0)$ for possible inclusion in the range or to rule out. However nothings strikes immediately except perhaps "tedius calculations", so will update in case something occurs, or perhaps someone else has a better idea...

• Thanks. I did the same calculation. I really want to know a 'special' way. – mathlove Sep 16 '13 at 7:01