If Rolle's theorem holds for the function $f(x)=x^{3}-a x^{2}+b x-4, x \in[1,2]$ with $f ^{\prime}\left(\frac{4}{3}\right)=0,$ then ordered pair $( a , b )$ is equal 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:

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)

$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$

$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$

$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$

$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is

The value of $c$ in the Lagrange's mean value theorem for the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}-4 \mathrm{x}^{2}+8 \mathrm{x}+11$ when $\mathrm{x} \in[0,1]$ is

Examine the applicability of Mean Value Theorem:

$(i)$ $f(x)=[x]$ for $x \in[5,9]$

$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$

$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$

Functions $f(x)$ and $g(x)$ are such that $f(x) + \int\limits_0^x {g(t)dt = 2\,\sin \,x\, - \,\frac{\pi }{2}} $ and $f'(x).g (x) = cos^2\,x$ , then number of solution $(s)$ of equation $f(x) + g(x) = 0$ in $(0,3 \pi$) is-