If $x = \sin(t), y = \cos(pt)$, then $(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = $

If the function $f(x) = ax^3 + bx^2 + 11x - 6$, defined on $[1, 3]$, satisfies all the conditions of Rolle's theorem for $c = 2 + \dfrac{1}{\sqrt{3}}$, then

The tangent to the curve $y^2 - xy + 9 = 0$ is vertical when

The value of $k$ such that the function $f(x) = \sin x - \cos x - kx + b$ is strictly decreasing for all real $x$ is