Given:
\( e^x \sin x = 1 \) has two real roots → say \( x_1 \) and \( x_2 \)
Apply Rolle’s Theorem:
Since \( f(x) = e^x \sin x \) is continuous and differentiable, and \( f(x_1) = f(x_2) \), ⇒ There exists \( c \in (x_1, x_2) \) such that \( f'(c) = 0 \)
Compute:
\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]
\[ \boxed{\text{At least one root}} \]
Not Available
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
Online Test Series, Information About Examination,
Syllabus, Notification
and More.