If the function $f(x) = 2x^2 + 3x + 5$ satisfies the Lagrange's Mean Value Theorem $(LMVT)$ at $x = 3$ on the closed interval $[1, a]$,then the value of $a$ is equal to:

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

Let $f^{\prime}(0)=-3$ and $f^{\prime}(x) \leq 5$ for all real values of $x$. The $f(2)$ can have possible maximum value as

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

Let $f(x)$ and $g(x)$ be two differentiable functions in $R$ such that $f(2) = 8, g(2) = 0, f(4) = 10$,and $g(4) = 8$. Then which of the following is true?

Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1, 0$ and $2$ be as given in the following table:

$x$	$x=-1, 0, 2$
$f(x)$	$3, 6, 0$
$g(x)$	$0, 1, -1$

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3g)^{\prime \prime}$ never vanishes. Then the correct statement$(s)$ is(are):
$(A)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup (0,2)$
$(B)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$