Between any two real roots of the equation e x sin x = 1, the equation e x cos x = - 1 has Options: A . &nbsp Atleast one root B . &nbsp Exactly one root C . &nbsp Atmost one root D . &nbsp No root

Answer: Option A : A Let ∝ , β ( ∝ < β ) be any two real roots of f(x) = e - x - sin x Then, f ( ∝ ) = 0 = f ( β ) Moreover, f(x) is continuos and differentiable for x ε [ ∝ , β ] . Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝ , β such t f ′ ( x ) = 0 ⇒ − e − x − c o s x = 0 ⇒ − e − x ( 1 + e x c o s x ) = 0 ⇒ e x c o s x = − 1. Hence (a) is the correct answer.

Between any two real roots of the equation ex sin x = 1, the equation ex

Discussion Forum : Application Of Derivatives

Question -

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

Options:

A . Atleast one root

B . Exactly one root

C . Atmost one root

D . No root

Answer: Option A
:
A

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

Was this answer helpful ?

Next Question

Discussion Forum : Application Of Derivatives

Submit Your Solution hear: