Question

Between any two real roots of $e^{-x}-\cos x=0$, there exists atleast $k$ root(s) of $\sin x-e^{-x}=0.$ Then $k$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Let $f\left(x\right)=e^{-x}-\cos x$ and let $\alpha$ and $\beta$ be two of many roots of the equation
$e^{-x}-\cos x=0$
$\Rightarrow f\left(\alpha \right)=0$ and $f\left(\beta \right)=0$
Also $f\left(x\right)$ is continuous and differentiable.
Then, according to the Rolle's theorem, there exists at least one $c\in (\alpha ,\beta )$ such that $f^{\prime }\left(c\right)=0$
$f^{\prime }\left(x\right)=-e^{-x}+\sin x$
$f^{\prime }\left(c\right)=-e^{-c}+\sin c=0$
$\Rightarrow c$ is root of the equation $e^{-x}-\sin x=0$