Question

Between any two real roots of the equation $e^x\sin x=1$, the equation $e^x\cos x=-1$ has ?

• A. Atleast one root
• B. Exactly one root
• C. Atmost one root
• D. No root

Answer 1

The correct option is A Atleast one root

$e^x\sin x=1\Rightarrow \sin x=e^{-x}$

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

let $f\left(x\right)=0$ at $x=a$ and $b(a<b)$ then there exist at least one c, for which

$f^{\prime }\left(c\right)=0\forall a<c<b$

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

at $x=c$

$\cos c+e^{-c}=0$

$\cos c=-e^{-c}$

$e^c\cos c=-1$

So for $x=c,g\left(x\right)=e^x\cos x+1$ become zero