Let $g(x) = f(x) + f(1-x)$ and $f''(x) < 0, 0 \leq x \leq 1$, then $\ldots$

If the function $f(x) = ax^2 + bx + \sin x$ satisfies all the conditions of Rolle's theorem on $[0, \pi]$ and the slope of the tangent to the curve $y = f(x)$ at $x = \dfrac{\pi}{4}$ is zero, then $a - b = $

If a particle moves such that the displacement (s) is proportional to the square of the velocity (v), then its acceleration (a) is

If $\int e^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{\sec^2(\tan^{-1}x)}\right)dx = e^{f(x)} + c$, then $\ldots$