Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then
$f(x) = 0$ has no root in $(0,1)$
$f(x) = 0$ has at least one root in $(0,1)$
$f' (c)$ vanishes for some $c\, \in \,(0,1)$
Both $(B)$ and $(C)$
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
If the equation
${a_n}{x^{n - 1}} + \,{a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$
has a positive root $x= \alpha ,$ then the equation
$n{a_n}{x^{n - 1}} + \,(n - 1){a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1} = 0$
has a positive root which is
Let $f(x)$ satisfy all the conditions of mean value theorem in $[0, 2]. $ If $ f (0) = 0 $ and $|f'(x)|\, \le {1 \over 2}$ for all $x$ in $[0, 2]$ then
In $[0, 1]$ Lagrange's mean value theorem is $ NOT$ applicable to
If Rolle's theorem holds for the function $f(x) = 2{x^3} + b{x^2} + cx,\,x\, \in \,\left[ { - 1,1} \right]$ at the point $x = \frac{1}{2}$ , then $(2b+c)$ is equal to