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
$3$
$4$
$5$
$1$
If $(1 -x + 2x^2)^n$ = $a_0 + a_1x + a_2x^2+..... a_{2n}x^{2n}$ , $n \in N$ , $x \in R$ and $a_0$ , $a_2$ and $a_1$ are in $A$ . $P$ .,then there exists
A value of $c$ for which conclusion of Mean Value Theorem holds for the function $f\left( x \right) = \log x$ on the interval $[1,3]$ is
Mean value theorem $f(b) -f(a) = (b -a) f '(x_1);$ from $a < x_1 < b,$ if $f(x) = 1/x$ then $x_1 = ?$
In the mean value theorem, $f(b) - f(a) = (b - a)f'(c)$if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$ is
The value of $\left[ {\frac{{\log \left( {\frac{x}{e}} \right)}}{{x - \,e}}} \right]\,\forall x\, > \,e$ is equal to (where [.] denotes greatest integer function)