In the mean value theorem, $f(b) - f(a) = (b - a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$ is
$8$
$5.25$
$4$
$6.25$
Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is....................
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 $(0,2)$
For every pair of continuous functions $f, g:[0,1] \rightarrow R$ such that $\max \{f(x): x \in[0,1]\}=\max \{g(x): x \in[0,1]\}$, the correct statement$(s)$ is (are) :
$(A)$ $(f(c))^2+3 f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$
$(B)$ $(f(c))^2+f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$
$(C)$ $(f(c))^2+3 f(c)=(g(c))^2+g(c)$ for some $c \in[0,1]$
$(D)$ $(f(c))^2=(g(c))^2$ for some $c \in[0,1]$
Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then, the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is
If $f(x)$ satisfies the conditions of Rolle’s theorem in $[1,\,2]$ and $f(x)$ is continuous in $[1,\,2]$ then $\int_1^2 {f'(x)dx} $ is equal to