Question

Between any two roots of $e^x\cos x=1$, there exists at least one root of $e^x\sin x=1$

• A. True
• B. False

Answer 1

The correct option is A True

Given the $e^x\cos x=1$

Rearranging we have $\cos x=e^{-x}$

$f\left(x\right)=\cos x-e^{-x}$

Assume that $a$ and $b$ are the roots of $f$

Then $f\left(a\right)=f\left(b\right)-0$

$f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Hence by Roll's theorem $c\in (a,b)\Rightarrow -\sin x+e^{-x}$

$i.e-\sin c^{-e^{-c}}=0$

$e^c\sin C-1=0$