If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is
${{35} \over {16}}$
${{35} \over {48}}$
${7 \over {16}}$
${5 \over {16}}$
Which of the following function can satisfy Rolle's theorem ?
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
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 $(0,2)$
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 $f$ is a differentiable function such that $f(2x + 1) = f(1 -2x)$ $\forall \,\,x \in R$ then minimum number of roots of the equation $f'(x) = 0$ in $x \in \left( { - 5,10} \right)$ ,given that $f(2) = f(5) = f(10)$ , is