A differentiable function satisfies $f(x) = \int_0^x [f(t)\cos t - \cos(t-x)]dt$. which is of the following hold good?
Step-by-Step Solution
Key Concept: Differentiate the integral equation to convert it into a standard linear first-order ODE, then solve using an integrating factor.
Given $f(x) = \int_0^x f(t)\cos t - \cos(t-x)dt$, we rewrite using substitution to get $f(x) = \int_0^x f(t)\cos t dt - \sin x$. Differentiating both sides yields $f'(x) = f(x)\cos x - \cos x$. Setting $y = f(x)$ and $\frac{dy}{dx} = f'(x)$, we solve the linear ODE $\frac{dy}{dx} - y\cos x = -\cos x$ with integrating factor $e^{-\sin x}$. This gives $ye^{-\sin x} = Ce^{\sin x} + 1$. Using the boundary condition at $x=0$: $y=0$, we find $C=-1$. Therefore, $f(x) = 1 - e^{\sin x}$. The minimum value is $1-e$ (when $x=\pi/2$) and maximum value is $1-e^{-1}$ (when $x=-\pi/2$).
Correct Answer: 1,2,3