If function $f(x) = x(x + 3) e^{-x/2} ;$ satisfies the rolle's theorem in the interval $[-3, 0],$ then find $C$

If $f: \mathrm{R} \rightarrow \mathrm{R}$ is a twice differentiable function such that $f^{\prime \prime}(x)>0$ for all $x \in \mathrm{R}$, and $f\left(\frac{1}{2}\right)=\frac{1}{2}, f(1)=1$, then

Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -

Consider the function $f (x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the mean value theorem, is

If the function $f(x) = a{x^3} + b{x^2} + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

If Rolle's theorem holds for the function $f(x) = 2{x^3} + b{x^2} + cx,\,x\, \in \,\left[ { - 1,1} \right]$ at the point $x = \frac{1}{2}$ , then $(2b+c)$ is equal to