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Find the number of real solutions of the equation $$\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1\,.$$

By hit and trial i got the solution at $x=\pm 1$ but i am not able to solve it as it involves power of 7

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Note that $1+\cos(\pi x)\geq 0$ for all $x\in\mathbb{R}$. Therefore, $$1-\frac{7^{1+\cos(\pi x)}}{3}\leq 1-\frac{1}{3}=\frac{2}{3}\,.$$ Thus, if $x$ is a real solution of the required equation, then $$3^{x^2-1}+3^{2\big(1-|x|\big)}=3\,\left(3^{x^2-2}+9^{\frac12-|x|}\right)=3\,\left(1-\frac{7^{1+\cos(\pi x)}}{3}\right)\leq 2\,.$$ By the AM-GM Inequality, $$3^{x^2-1}+3^{2\big(1-|x|\big)}\geq 2\,\sqrt{3^{x^2-1}\cdot3^{2\big(1-|x|\big)}}=2\cdot3^{\frac{\big(|x|-1\big)^2}{2}}\geq 2\,.$$ Therefore, the inequality above must be an equality, which means $x^2-1=2\big(1-|x|\big)$ and $\big(|x|-1\big)^2=0$. This shows that $|x|=1$, or $x=\pm1$.

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  • $\begingroup$ Ingenious! +1! =) $\endgroup$ Jul 1 '20 at 11:41
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Note that $\frac{7^{1+cos\pi x}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}$ is an even function. So it's enough to consider $x\ge 0$.

Let $$f(x) = \frac{7^{1+cos\pi x}}{3}+3^{x^2-2}+9^{\frac{1}{2}-x}$$ If $x \gt \sqrt{2} \ $ then $3^{x^2 - 2}\gt 1$ and also $\frac{7^{1+cos\pi x}}{3} \gt 0 \ \ , \ 9^{\frac{1}{2}-x} \gt 0$. Therefore $f(x) \gt 1$ and there is no root in this case. In the interval $0\le x \le 1$ we will show $\frac{7^{1+cos\pi x}}{3}$ is a decreasing function and also $3^{x^2-2}+9^{\frac{1}{2}-x}$. So there sum will be decreasing and the only root occurs when $x = 1$. When $1\lt x \le \sqrt{2}$ we have to compute the derivative $$f'(x) = -\frac{7\pi\ln 7}{3}\times7^{\cos \pi x} \sin{\pi x} + (2x)(\ln 3) 3^{x^2 -2} + (-2)(\ln 3)3^{1-2x}$$ In the mentioned interval $-\sin \pi x$ and $\cos \pi x$ are increasing so $-\frac{7\pi\ln 7}{3}\times7^{\cos \pi x} \sin{\pi x}$ is increasing. In the similar manner we can show two other functions are increasing, so $f'(x)$ is increasing when $1\lt x \le \sqrt{2}$. So we have $f'(x) \gt f'(1) = 0$. We can conclude there is no root in $1\lt x \le \sqrt{2}$.

Let $0\le x \le 1$ then $$g(x) = \frac{7^{1+cos\pi x}}{3} \implies g'(x) = -\frac{7\pi\ln 7}{3}\times7^{\cos \pi x} \sin{\pi x} \le 0$$ And $$h(x) = 3^{x^2-2}+9^{\frac{1}{2}-x} \implies h'(x) = (2x)(\ln 3) 3^{x^2 -2} + (-2)(\ln 3)3^{1-2x} = 2\ln 3(x\times 3^{x^2 - 2} - 3^{1 - 2x}) \le 0$$ Because $$x\times 3^{x^2 - 2} - 3^{1 - 2x} \le 0 \iff \log_{3} (x\times 3^{x^2 - 2}) \le \log_{3} (3^{1 - 2x}) \iff \log_{3} (x) + x^2 - 2 \le 1 -2x \iff \log_{3} (x) \le -x^2 -2x + 3 \le 3$$ Clearly this answer shows how much Batominovski's solution is elegant!

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