If $f'(x) = \sin^2 x$ and $y = f\left(\dfrac{2x-1}{x^2+1}\right)$, then $\dfrac{dy}{dx}$ at $x = 1$ is

If $f : R \to R$ is an even function then

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 = $