Suppose that $f(0) = -3$ and $f'(x) \le 5$ for all values of $x$. Then the largest value which $f(2)$ can attain is

Let $f$ be a function that is continuous and differentiable for all real $x$. If $f(2) = -4$ and $f'(x) \geq 6$ for all $x \in [2, 4]$,then which of the following is true?

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

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 $(0,2)$

Let $f$ be continuous on $[1, 5]$ and differentiable in $(1, 5).$ If $f(1)=-3$ and $f'(x) \ge 9$ for all $x \in (1, 5)$,then which of the following is true?

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f(0) = 2$,$g(1) = 2$,$g(0) = 0$,and $f(1) = 6$,then for some $c \in (0, 1)$: