Answer to Question #103851 in Calculus for BIVEK SAH

Question

Prove that between any two real roots of the equation
2e^2xsin3x-3=0

There is at least one real root of the equation
e^2xcos3x+1=0

Answer 1

We consider the following function:

$f(x)=\frac{1}{3}\sin 3x -\frac{1}2{}e^{-2x}$

$\sin 3x$ and $e^{-2x}$ are continuous and differentiable for all real $x$ $\Rightarrow \ f(x)$ is continuous and differentiable for all real $x.$

Suppose, that $a$ and $b$ ($a<b)$ are real roots of $2e^{2x}\sin 3x-3=0$ .

$2e^{2x}\sin 3x-3=6e^{2x}(\frac{1}{3}\sin3x-\frac{1}{2} e^{-2x})=0$

$e^{2x}$ ≠$\ 0 \Rightarrow \frac{1}{3}\sin3x-\frac{1}{2}e^{-2x}=0.$

So, if $a$ and $b$ are roots of $2e^{2x}\sin 3x-3=0$ , then $a$ and $b$ are roots of $f(x)=0$ ,

i.e. $f(a)=f(b)=0.$

Now we can use Rolle’s Theorem:

$f(x)$ is continuous in $[a,b]$ , differentiable in $(a,b)$ , $\ f(a)=f(b)=0.$

Therefore, $\exists \ c\in(a,b): \ f^\prime (c)=0$ .

$f^\prime (c)=\cos(3c)+e^{-2c}=0 \ \Rightarrow e^{2c}\cos3c+1=0$

So, for all real roots ($a$ and $b)$ of $2e^{2x}\sin 3x-3=0$ , there is $c\in(a,b) : e^{2c}\cos3c+1=0$ .