0
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ "maximum value of the function is when denominator term is minimum " of which function? $\endgroup$
    – leonbloy
    Oct 24 '14 at 18:01
  • $\begingroup$ Maybe you are just supposed to graph it? $\endgroup$
    – Jacob Bond
    Oct 24 '14 at 18:12
0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ $\sin$ 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 $[\cos1,1)$. $\endgroup$ Oct 24 '14 at 18:36
  • $\begingroup$ I edited the post to contain this. $\endgroup$
    – hjhjhj57
    Oct 24 '14 at 18:49
  • 1
    $\begingroup$ 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 $[a_1+b_1,a_2+b_2]$.) $\endgroup$ Oct 24 '14 at 19:48
  • $\begingroup$ I'll be quite busy for the next couple of days, but feel free to edit my answer :) $\endgroup$
    – hjhjhj57
    Oct 24 '14 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.