If $f(x)$ is a differentiable function,$f^{\prime}(x) \geq 5$ for all $x \in [2, 6]$,$f(2) = 4$ and $f(3) = 15$,then a possible value of $f(6)$ is:

Let $f(x)$ be a function continuous on $[1, 2]$ and differentiable on $(1, 2)$ satisfying $f(1) = 2, f(2) = 3$ and $f'(x) \geq 1$ for all $x \in (1, 2)$. Define $g(x) = \int_1^x f(t) \, dt$ for all $x \in [1, 2]$. Then the greatest value of $g(x)$ on $[1, 2]$ is-

For which real value of $K$ does the equation $2x^3 + 3x + K = 0$ have two real roots in the interval $[0, 1]$?

If $f(x)=(2x-1)(3x+2)(4x-3)$ is a real-valued function defined on $[\frac{1}{2}, \frac{3}{4}]$, then the value$(s)$ of '$c$' as defined in the statement of Rolle's theorem is/are:

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 $(0,2)$

The function $f(x) = (x - 3)^2$ satisfies all the conditions of the Mean Value Theorem in $[3, 4]$. $A$ point on $y = (x - 3)^2$,where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$,is