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

Let $\alpha ,\beta (\alpha <\beta )$ be any two real roots of

$f\left(x\right)=e^{-x}-\sin x$ Then, $f\left(\alpha \right)=0=f\left(\beta \right)$

More over, $f\left(x\right)$ is continuous & differentiable

for $x\in [\alpha ,\beta ]$. From Rolle's theorem, there

exists atleast one $x$ in $(\alpha ,\beta )$ such that

$f^{\prime }\left(x\right)=0$

$\Rightarrow -e^{-x}-\cos x=0$

$\Rightarrow -e^{-x}(1+e^x\cos x)=0\Rightarrow e^x\cos x=-1$