Q. If $\phi(x) = \cos{x - \int_{0}^{x}{(x - t)\ \phi(t)dt,\ }}$ then $\phi^{''}(x) + \phi(x)$ is equal to
A) $- \cos x$
B) 0
C) $\int_{0}^{x}{(x - t)\phi(t)dt}$
D) $- \int_{0}^{- x}{(x - t)\phi(t)dt}$
Explanation
(a)
$\phi(x) = \cos{x - \int_{0}^{x}{(x - t)\phi(t)dt}}$
$= \cos{x - x\int_{0}^{x}{\phi\ (t)dt + \int_{0}^{x}{t\ \phi(t)dt}}}$
$\Rightarrow \ \phi^{'}(x) = - \sin{x -}\int_{0}^{x}{\phi(t)dt} - x\phi(x) + x\phi(x)$
$\Rightarrow \ \phi^{''}(x) = - \cos{x - \phi(x)}$
$\Rightarrow \ \phi(x) + \phi^{''}(x) = - \cos x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.