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Solve the equation $$2^{1-x} + 2^{\sqrt{2x-x^2}}=3 \tag 1$$ on reals, using elementary knowledge (using trigonometry or logarithms is allowed, but without limits, differential calculus etc.)

We have to find solutions on $[0,2]$ interval. Two solutions are easy to spot $x=0, x=1$

By rewriting (1) we get: $$2^{1-x} =3-2^{\sqrt{2x-x^2}} \tag 2$$ The left side is decreasing. The right side is decreasing on $[0,1]$ and increasing on $[1,2]$. It follows that there are no solutions in $(1, 2]$.

The only interval I cannot cover is (0, 1).

Any help is appreciated.

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    It can be transformed $2^{siny}+2^{cosy}=3$ – Takahiro Waki Jun 11 '16 at 11:23
  • @TakahiroWaki Might be a good idea, but I can't see how to use it. –  Jun 11 '16 at 11:27
  • $x=0$ is the only solution – Archis Welankar Jun 11 '16 at 11:30
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    @ArchisWelankar $x=1$ is also solution –  Jun 11 '16 at 11:31
  • yes I missed that – Archis Welankar Jun 11 '16 at 11:33
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    What is essential to this problem (at least in my opinion) is that $\sqrt{2x-x^2}$ grows faster than the rate at which $1-x$ decreases, so $-2^\sqrt{2x-x^2}$ decreases faster than $2^{1-x}$ on the interval $(0, 1)$ and thus there are no solutions on that interval. I don't know how to show any of this rigorously, though, unless we get into calculus. – Noble Mushtak Jun 11 '16 at 12:08
  • I think there are only two solution x=0 and 1 – Aakash Kumar Jun 11 '16 at 12:15
  • @user346279 That's what I said in my comment above. – Noble Mushtak Jun 11 '16 at 12:15
  • You can show it by converting into sinx abd cosx – Aakash Kumar Jun 11 '16 at 12:18
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    You are right, but I think it still needs calculus. If we use @TakahiroWaki 's idea and analyze $f(y)=2^{\sin y}+2^{\cos y}$. There is a maximum at $y=\frac \pi 4$, where $f(y) > 3$ and a minimum at $\frac{5\pi}{4}$, where $f(y) < 3$. Thus, there is an intersection on the increasing interval $[0, \frac \pi 4]$ at $y=0$, another intersection on the decreasing interval $[\frac \pi 4, \frac{5\pi}{4}]$ at $y=\frac \pi 2$, and then no intersection on the increasing interval $[\frac{5\pi}{4}, 2\pi)$ because $f(2\pi)=3$ so $f(y) < 3$ on this interval. Thus, there are only two intersections. – Noble Mushtak Jun 11 '16 at 12:18
  • @TakahiroWaki, I made your comment work. See my answer. – Barry Cipra Jun 11 '16 at 16:12
  • Not sure sin, cos works. If 1-x = sin, then $\sqrt{2x - x^2} = \sqrt{(1-x)^2 - 1} \ne \sqrt{1 - (1-x)^2} = \cos$. – fleablood Jun 11 '16 at 17:36
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    $\sqrt{2x-x^2}≠\sqrt{(1-x)^2-1}$ – Takahiro Waki Jun 12 '16 at 00:41
  • @fleablood, plugging $x=1$ into your comment, you get $\sqrt1=\sqrt{-1}$.... – Barry Cipra Jun 13 '16 at 14:42

2 Answers2

1

To begin with a seemingly random remark, note that

$$4(1-s^2)\gt(2-s)^2\quad\text{for }0\lt s\le{1\over\sqrt2}\approx0.707$$

This is because the inequality simplifies to $s(4-5s)\gt0$, which holds for $0\lt s\lt{4\over5}=0.8$.

Now down to business. As noted by Takahiro Waki in comments, the equation $2^{1-x}+2^{\sqrt{2x-x^2}}=3$ is equivalent to

$$2^{\sin\theta}+2^{\cos\theta}=3$$

with $-{\pi\over2}\le\theta\le{\pi\over2}$. It's easy to see that $2^{\sin\theta}+2^{\cos\theta}\lt3$ for $-{\pi\over2}\le\theta\lt0$, and equality is achieved at $\theta=0$ and $\theta={\pi\over2}$ (corresponding to $x=1$ and $x=0$, respectively). So it remains to show

$$2^{\sin\theta}+2^{\cos\theta}\gt3\quad\text{for }0\lt\theta\lt{\pi\over2}$$

Actually, by the symmetry $\sin({\pi\over2}-\theta)=\cos\theta$ (and vice versa), we need only prove it in the interval $0\lt\theta\le{\pi\over4}$. Now let's abbreviate this to

$$2^s+2^c\gt3$$

where $0\lt s\le{1\over\sqrt2}$ and $c=\sqrt{1-s^2}$. To prove this, let's make clever use of AGM:

$$2^s+2^c=2^s+2^{c-1}+2^{c-1}\ge3\sqrt[3]{2^{s+2c-2}}$$

Thus we need only prove $s+2c-2\gt0$ for $0\lt s\le{1\over\sqrt2}$. But this is straightforward: Since $c\ge0$, we have

$$\begin{align} s+2c-2\gt0 &\iff2c\gt(2-s)\\ &\iff4c^2\gt(2-s)^2\\ &\iff4(1-s^2)\gt(2-s)^2 \end{align}$$

which takes us back to the obviously now non-random remark at the beginning.

Barry Cipra
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  • Firstly, I thought what first inequality is. My computation is $2^{siny}+2^{cosy}\geq 2^{(siny+cosy)/2+1}\geq 2^{\sqrt2sin(y+pi/4)/2+1}\geq 2.8$,but this RHD is minimum at y=0. Finally, LHD$\geq 3$ – Takahiro Waki Jun 12 '16 at 00:59
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The function $f(x)=2^{1-x} + 2^{\sqrt{2x-x^2}}$ has as domain $[0,2]$ and the derivative is $f'(x)= \log 2(\frac{2^{\sqrt{2x-x^2}}(x-1)}{\sqrt{2x-x^2}}-2^{1-x})$.

$f$ has a maximun at $x_m$, $f(x_m)\gt 3$ and $f(2)=\frac 32$, furthermore $f$ is decreasing in the interval $[x_m,2]$.

It follows since $f(0)=3$ there is an unique other value in the domain $[0,2]$ for which $f(x)=3$. This value is apparent and equal to $1$.

Thus the only soltions are $\color{red}0$ and $\color{red}1$.

Piquito
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