# Find the range of function $f(x) =\cos\sin\ln\left(\frac{x^2+e}{x^2+1} \right)+\sin\cos\ln\left(\frac{x^2+e}{x^2+1} \right)$

Problem :

Find the range of function $$f\left( x \right) =\cos \left( \sin \left( \ln \left( \frac{x^2+e}{x^2+1} \right) \right) \right) +\sin \left( \cos \left( \ln \left( \frac{x^2+e}{x^2+1} \right) \right) \right)$$

My approach :

maximum value of the function is when denominator term is minimum i.e. $$x^2+1$$ is minimum.

It is minimum when $$x^2 =0$$ therefore, maximum value of function $$\cos(\sin(\ln e))+\sin(\cos(\ln e))$$

$$\cos(\sin(1))+\sin(\cos(1))$$ now how to find the minimum value of the function please suggest . Thanks...

• "maximum value of the function is when denominator term is minimum " of which function? Oct 24, 2014 at 18:01
• Maybe you are just supposed to graph it? Oct 24, 2014 at 18:12

Well, first you need to see the range of $\frac{x^2+e}{x^2+1}$, which you can easily verify to be $(1,e]$. Then, the range of $\log$ in this domain is $(0,1]$.

As pointed out in a comment:

Since $\sin$ is increasing and $\cos$ decreasing in $(0,1]$, then $\sin(0,1] =(0,\sin(1)]$ and $\cos(0,1] =[\cos(1),1)$.

Finally, since in $(0,\sin(1)]$ the cosine is decreasing and in $[\cos(1),1)$ the sine is increasing, we get that the range of each part of the sum is $[\cos(\sin(1)),1)$ and $[\sin(\cos(1)),1)$. So the range of the complete function is $$[\cos(\sin(1))+\sin(\cos(1)), 2)$$

Another way could be using the fact that $\log(a/b) = \log a - \log b$, but this seems like a rather cumbersome approach.

• $\sin$ is increasing on $-\pi/2<x<\pi/2$, so the $\sin$ of $(0,1]$ has range $(0,\sin1]$. Similarly for $\cos$, which is decreasing on $0<x<\pi$, so the $\cos$ of $(0,1]$ has range $[\cos1,1)$. Oct 24, 2014 at 18:36
• I edited the post to contain this. Oct 24, 2014 at 18:49
• Just because the range of $f(x)$ is $[a_1,a_2]$ and of $g(x)$ is $[b_1,b_2]$ does not necessarily mean that the range of $f(x)+g(x)$ is $[a_1+b_1,a_2+b_2]$. Example: Let $f(x)=-\sin x,g(x)=\sin x$. (Of course, it must be a subset of $[a_1+b_1,a_2+b_2]$.) Oct 24, 2014 at 19:48
• I'll be quite busy for the next couple of days, but feel free to edit my answer :) Oct 24, 2014 at 22:30