Question

The equation $e^{\sin x}-e^{-\sin x}-4=0$ has

• A. infinite number of real roots
• B. no real roots
• C. exactly one real root
• D. exactly four roots

Answer 1

The correct option is B

no real roots

Explanation for the correct option:

Finding the roots for the given equation:

Given

$\begin{array}{rcl}{e}^{\sin x}-e^{-\sin x}-4& =& 0\\ Let{e}^{\sin x}& =& p\\ p-\frac{1}{p}-4& =& 0\\ p^2-1-4p& =& 0\\ p^2-4p-1& =& 0\end{array}$

Finding the roots of the equation by using the formula $\begin{array}{l}\mathbf{x}\mathbf{}\mathbf{=}\frac{\mathbf{}\mathbf{(}\mathbf{-}\mathbf{b}\mathbf{}\mathbf{\pm }\mathbf{}\sqrt{\mathbf{(}\mathbf{b}\mathbf{²}\mathbf{}\mathbf{-}\mathbf{}\mathbf{4}\mathbf{a}\mathbf{c}\mathbf{)}}\mathbf{}\mathbf{)}}{\mathbf{2}\mathbf{a}}\end{array}$

$$\begin{gathered} \begin{array}{c}p=\frac{4\pm \sqrt{(4^2-4\left(-1\right))}}{2}\end{array} \\ \begin{array}{c}=\frac{4\pm \sqrt{16+4}}{2}\end{array} \\ \begin{array}{l}=\frac{4\pm \sqrt{20}}{2}\end{array} \\ \begin{array}{l}=\frac{4\pm \sqrt{4\times 5}}{2}\end{array} \\ \begin{array}{l}=\frac{4\pm 2\sqrt{5}}{2}\end{array} \\ \begin{array}{c}=\frac{2\left(2\pm 2\sqrt{5}\right)}{2}\end{array} \\ p=2\pm 2\sqrt{5} \end{gathered}$$

Since $p$ is a real positive number, we can't consider the root $\begin{array}{rcl}p& =& 2-\sqrt{5}\end{array}$

$\begin{array}{rcl}{e}^{\sin x}& =& 2+\sqrt{5}\\ \sin x& =& {\log }_e\left(2+\sqrt{5}\right)\end{array}$

But $\begin{array}{rcl}2+\sqrt{5}& >& e\end{array}$, so applying $\log$ on both sides we get

$\begin{array}{rcl}{\log }_e\left(2+\sqrt{5}\right)& >& {\log }_e\left(e\right)\\ {\log }_e\left(2+\sqrt{5}\right)& >& 1\end{array}$

Here LHS is greater than $1$,which is equal to $\sin x$.

Then,$\sin \left(x\right)>1$is not possible for all the values of $x$

So,the above equation does not have any real solution.

Hence, option (B) is the correct answer.