If the function $f(x) = 2x^2 + 3x + 5$ satisfies $LMVT$ at $x = 3$ on the closed interval $[1, a]$ then the value of $a$ is equal to
$3$
$4$
$5$
$1$
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 $(0,2)$
Verify Rolle's theorem for the function $y=x^{2}+2, a=-2$ and $b=2$
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
If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ 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. . . . . . . .