If function $f(x) = x(x + 3) e^{-x/2} ;$ satisfies the rolle's theorem in the interval $[-3, 0],$ then find $C$
$0$
$1$
$-2$
$-3$
Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2}\ln x,\,x > 0} \\
{0,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0}
\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $
If the function $f(x) = 2x^2 + 3x + 5$ satisfies $LMVT$ at $x = 3$ on the closed interval $[1, a]$ then the value of $a$ is equal to
Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then
The function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfy the conditions of Rolle's theorem in $[1, 3]. $ The values of $a $ and $ b $ are
If the Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1 ]$ for the point $c = \frac{1}{2}$ , then the value of $2a + b$ is