If the function $f(x) = a{x^3} + b{x^2} + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

  • A

    $1, -6$

  • B

    $-2, 1$

  • C

    $-1$, $\frac{1}{2}$

  • D

    $-1, 6$

Similar Questions

Verify Rolle's Theorem for the function $f(x)=x^{2}+2 x-8, x \in[-4,2]$

 $(i)$  $f (x)$ is continuous and defined for all real numbers

$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4)  = 0$

$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$

$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.

$(v)$ the signs of  $f '(x)$ is given below

Possible graph of $y = f (x)$ is

For a polynomial $g ( x )$ with real coefficient, let $m _{ g }$ denote the number of distinct real roots of $g ( x )$. Suppose $S$ is the set of polynomials with real coefficient defined by

$S=\left\{\left(x^2-1\right)^2\left(a_0+a_1 x+a_2 x^2+a_3 x^3\right): a_0, a_1, a_2, a_3 \in R\right\} \text {. }$

For a polynomial $f$, let $f^{\prime}$ and $f^{\prime \prime}$ denote its first and second order derivatives, respectively. Then the minimum possible value of $\left(m_f+m_{f^{\prime}}\right)$, where $f \in S$, is. . . . . . . .

  • [IIT 2020]

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:

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

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 $(0,2)$

  • [IIT 2015]

Given $f (x) =4\,\, - \,\,{\left( {\frac{1}{2}\, - \,x} \right)^{2/3}}\,$ $g (x) = \left\{ \begin{array}{l}\frac{{\tan \,\,[x]}}{x}\,\,\,\,,\,\,x \ne \,0\\1\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x\, = \,0\end{array} \right.$

$h (x) = \{x\}$   $k (x) = {5^{{{\log }_2}(x\, + \,3)}}$then in $[0, 1]$ Lagranges Mean Value Theorem is $NOT$ applicable to