$e^{|\sin x|}+e^{-|\sin x|}+4a=0$ will have exactly four different solution in $[0,2\pi]$

Question

$e^{|\sin x|}+e^{-|\sin x|}+4a=0$ will have exactly four different solution in $[0,2\pi]$ if

(A) $a\in\bigg[\frac{-e}{4},\frac{-1}{4}\bigg]$

(B) $a\in\bigg[\frac{-1-e^2}{4e},\infty\bigg)$

(C) $a\in \mathbb R$

(D) None of these

My Approach:

Let $e^{|\sin x|}=t \in[1,e]$

So equation transformed into $t^2+4at+1=0$

Above equation must have two distinct solution in $[1,e]$

For two distinct solution we must have $f(1)\geq0$ , $f(e) \geq0$ , $4a^2-4 \gt0$ and $1\lt -2a \lt e$

where $f(t) = t^2+4at+1$

Is my Approach Correct?

Answer 1

The number of $x$ satisfying $|\sin x|=s$ and $0\leqslant x\leqslant 2\pi$ is $$\begin{cases}3&\text{if $\ s=0$} \\4&\text{if $\ s\in (0,1)$} \\2&\text{if $\ s=1$} \\0&\text{if $\ s\in(-\infty,0)\cup(1,\infty)$}\end{cases}$$ (see the graph of $y=|\sin x|\ (0\leqslant x\leqslant 2\pi)$.)

So, the number of $x$ satisfying $e^{|\sin x|}=t$ and $0\leqslant x\leqslant 2\pi$ is $$\begin{cases}3&\text{if $\ t=1$} \\4&\text{if $\ t\in (1,e)$} \\2&\text{if $\ t=e$} \\0&\text{if $\ t\in(-\infty,1)\cup(e,\infty)$}\end{cases}$$

Let $t=e^{|\sin x|}$. Then, as you wrote, our equation can be written as

$$t^2+4at+1=0\tag1$$

Suppose that $a\geqslant 0$. Then, since $t\gt 0$, LHS of $(1)$ is positive. This is a contradiction. So, $a\lt 0$.

Also, considering the discriminant, one has $(4a)^2-4\geqslant 0$, i.e. $a\in \bigg(-\infty,-\dfrac 12\bigg]\cup \bigg[\dfrac 12,\infty\bigg)$.

Therefore, it is necessary that $a\leqslant -\dfrac 12$.

Let $p=-2a-\sqrt{4a^2-1},q=-2a+\sqrt{4a^2-1}$ be the solutions of $(1)$.

• Case 1 : $a=-\dfrac 12$
  
  Then, $p=q=1$, so $(1)$ has exactly three different solutions in $[0,2\pi]$.

• Case 2 : $a\lt -\dfrac 12$
  
  Then, since $p\lt 1\lt q$ (because LHS of $(1)$ is negative when $t=1$), $(1)$ has exactly four different solutions in $[0,2\pi]$ if and only if $$\begin{align}q\lt e&\iff -2a+\sqrt{4a^2-1}\lt e \\\\&\iff \sqrt{4a^2-1}\lt e+2a \\\\&\iff e+2a\gt 0\quad\text{and}\quad 4a^2-1\lt (e+2a)^2 \\\\&\iff -\frac{e}{2}\lt a\quad\text{and}\quad \frac{-e^2-1}{4e}\lt a \\\\&\iff \dfrac{-1 - e^2}{4 e}<a\ \ \bigg(\lt -\frac 12\bigg)\end{align}$$

As a result, $(1)$ has exactly four different solutions in $[0,2\pi]$ if and only if $$a\in \bigg(\frac{-1 - e^2}{4 e},-\frac 12\bigg)$$

Therefore, the correct answer is $\color{red}{(D)}$.

Answer 2

Say $s=|\sin x|$, then $s\in [0,1]$ and we need to find for which $s$ has $-4a = e^s+e^{-s}$ two solutions. Since function $f(s) = e^s+e^{-s}$ is even and increases for $s\geq 0$ we see that $f_{\max} = f(1) = e+{1\over e}$and $f_{\min} = f(0) = 2$. So $$2<-4a\leq e+{1\over e}$$ or $$-{1\over 2}>a\geq -{1+e^2\over 4e}$$ Since $-{1\over 4}>-{1\over 2}$ the answer is (D).