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The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$.

I tried to find the minimum and maximum value of the function.First i simplified the function.

$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{1+4\sin^2\left(\frac{x}{2}+\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)}$

Then i differentiated the function and equate it to zero to get the critical points.

Critical point equations are $\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0$

$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{2},\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{-1}{2}$

When i checked plotted the function on desmos.com graphing calculator,i found minimum value to be $0.5$ and maximum value to be $2.5$.

Which i cannot get by my critical points.Where have i gone wrong?Please help me.

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  • $\begingroup$ An approach : What is the minimum value, what is the maximum value (not local minima and maxima). Is the function continuous ? . Then will it not take all values between the 2 extremes ?. You have already simplified the function $\endgroup$ – Shailesh Aug 6 '15 at 11:32
  • $\begingroup$ Do you mean "contains precisely $N$ integers"? $\endgroup$ – Matemáticos Chibchas Sep 10 '15 at 3:10
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Put $\sqrt{1+\cos x}$ +$\sqrt{1-\cos x} = A$

$A^2 = 2\pm 2 \sin x ,\quad A^2 - 2 =\pm 2 \sin x$

$ -2\leq A^2 - 2\leq 2,\quad -2\leq A\leq2$

So $f(x) = \frac{5 - A^2}{A}$ or $\frac{A^2 + 1}{A}$

Find the minimum and maximum of $f(x)$ in the two conditions with $-2\leq A\leq 2$

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  • $\begingroup$ I'm not so sure about your approach for the following reason. When you say $\sqrt{1 + \cos x}$, I think we always mean the positive square root (I am open to correction in this respect). In that case, the denominator is always positive and so is the numerator. That's why I took 0.5 to 2.5 as the range. $\endgroup$ – Shailesh Sep 10 '15 at 14:09
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Hint:

The minimum value of the function is $1/2$ and the maximum is $2.5$. The function is clearly continuous. So it takes every value between these numbers, specifically 1 and 2. So $N=2$ which gives $10N$. Can you show that these are indeed the minimum and maximums. I have outlined the general approach

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  • $\begingroup$ How has minimum value come,Maximum value i got 2.5,but minimum i did not get. $\endgroup$ – Brahmagupta Aug 6 '15 at 11:56
  • $\begingroup$ The same way you got maximum $\endgroup$ – Shailesh Aug 6 '15 at 12:40
  • $\begingroup$ The denominator goes to 0 as $ x \rightarrow -\pi /2$ how can it be continuous there? $\endgroup$ – Subhasish Basak Sep 5 '15 at 0:25
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Your simplified equation is correct only for $0 \leq x \leq \pi$. During your derivation be sure to consider both positive and negative square roots in the denominator. The result is an alternate version of your simplified equation

$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{5-4\sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}-\frac{\pi}{4}\right)}$

valid for $\pi < x < 2\pi$. When you work this through you will get another critical point at $\cos(x/2 - \pi/4)=0$.

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