Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$

Question

Find the range of: $$y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$$

What I tried:

Let:$$\log_e\frac{x^2+e}{x^2+1}=X,$$ then $$y=\sqrt {\sin X}+\sqrt{\cos X}$$ $$y_{max}at X=\pi/4$$

The rest is too complicated. I am stuck up.

Can someone please give a analytical solution.

Even Wolfram alpha doesn't help.

Big hint: $\dfrac{x^2+e}{x^2+1} = 1 + \frac{e-1}{x^2+1}$. What can you say about the range of this function? Once you've worked that out work out the range of $X=\log \dfrac{x^2+e}{x^2+1}$ and then work out the range of $y=\sqrt{\sin X} + \sqrt{\cos X}$. — Tim, May 26 '13 at 11:32
@Tim: It is a good hint. By the way, I paid attention to your comment after I posted my answer. — Mhenni Benghorbal, May 26 '13 at 12:25

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

6

A related problem. First, we study the expression

$$ \frac{x^2+e}{x^2+1}=1+\frac{e-1}{x^2+1} \longrightarrow_{|x|\to\infty} 1, $$

which implies $ y(x)\longrightarrow_{|x|\to \infty} 0 $. To find the maximum of the function $y(x)$, Let's study the function

$$ h(t)=\sqrt{\sin(t)}+\sqrt{\cos(t)}. $$

The maximum of the above function is attained at $t=\frac{\pi}{4}$ which can proved using derivative test. So, this implies our function attains its max when

$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} \implies x=0.6632987771, -0.6632987771. $$

Plugging back in the function y(x) gives the max which is $y=1.681792830$. So the range is

$$ 1 < y \leq 1.681792830.$$

Note: You can solve $ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4} $ easily as,

$$ \ln\left( \frac{x^2+e}{x^2+1} \right)=\frac{\pi}{4}\implies \frac{x^2+e}{x^2+1} =e^{\frac{\pi}{4}}\implies x^2+e= e^{\frac{\pi}{4}}x^2+ e^{\frac{\pi}{4}}=\dots.$$

I think you can finish it.

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edited Apr 13 '17 at 12:21

Community

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answered May 26 '13 at 12:03

Mhenni Benghorbal

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I have the answer as $R_f=(1,2^{3/4}]$ – MathGeek May 26 '13 at 12:53
1.681792830~$2^{3/4}$ – MathGeek May 26 '13 at 12:54
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@MathGeek: See the note. You can get the exact value for $x$, where the function achieves its max, then substitute in the $y(x)$. – Mhenni Benghorbal May 26 '13 at 12:57
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@MathGeek: Yes, that's the range. It is a typo. – Mhenni Benghorbal May 26 '13 at 12:58
You are saying $0<y≤2^{3/4}$. But the answer is $y=(1,2^{3/4}]$ – MathGeek May 26 '13 at 12:59
@MathGeek: It is a typo. I already corrected it. – Mhenni Benghorbal May 26 '13 at 13:00
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Thanks. That was very helpful. and from where did you made the graph? – MathGeek May 26 '13 at 13:00
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@MathGeek: I used Maple 17. You can use Maple or Mathematica, both will give you the same answer, I believe. – Mhenni Benghorbal May 26 '13 at 13:01
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@MathGeek: You are welcome. Glad to assist. – Mhenni Benghorbal May 26 '13 at 13:02
let us continue this discussion in chat – MathGeek May 26 '13 at 15:19

score 0 · Answer 2 · answered May 26 '13 at 11:39

let $f(x)=\frac{x^2+e}{x^2+1} \to 1<f(x)\le e \to 0<X\le 1$

$y^2=sinX+cosX+\sqrt{2sin2X}=\sqrt{2}\left(sin(X+\dfrac{\pi}{4})+\sqrt{sin2X}\right)$

when $0<X<\dfrac{\pi}{4}$,$sin2X$ and $sin(X+\dfrac{\pi}{4})$ are mono increasing,$\to y^2>1 \to y>1$

I left another part for you to discuss and you should be able to find out the range now.

Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$

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