If $f(x)=\cos x-\int\limits_0^t(x-t)f(t)dt$, then f''(x) + f(x) equals:
Answer & explanation
Correct answer: option 4
$f(x)=\cos x-\int\limits_0^t(x-t)f(t)dt⇒f'(x)=-\sin x-\int\limits_0^t\frac{d}{dx}[(x-t)f(t)]dt$
$⇒f'(x)=-\sin x-\int\limits_0^t[f(t)-0]dt⇒f''(x)=-\cos x - 0$
$⇒f(x)+f''(x)=\cos x-\int\limits_0^t(x-t)f(t)dt+(-\cos x)$