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 $\propto ,\beta (\propto <\beta )$ be any two real roots of
f(x) = e - x - sin x
Then, f$(\propto )=0=f\left(\beta \right)$
Moreover, f(x) is continuos and differentiable for $x\epsilon [\propto ,\beta ].$
Hence, from Rolle's thereom, thereom, there exists atleast one x in $\propto ,\beta$ such t
$f^{\prime }\left(x\right)=0\Rightarrow -e^{-x}-cosx=0\Rightarrow -e^{-x}(1+e^{x}cosx)=0\Rightarrow e^{x}cosx=-1.$
Hence (a) is the correct answer.