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
$1 - {{\sqrt {15} } \over 6}$
$1 + \sqrt {15} $
$1 - {{\sqrt {21} } \over 6}$
$1 + \sqrt {21} $
From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$ $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $
Consider $f (x) = | 1 - x | \,;\,1 \le x \le 2 $ and $g (x) = f (x) + b sin\,\frac{\pi }{2}\,x$, $1 \le x \le 2$ then which of the following is correct ?
If from mean value theorem, $f'({x_1}) = {{f(b) - f(a)} \over {b - a}}$, then
Let $f(x)$ be a function continuous on $[1,2]$ and differentiable on $(1,2)$ satisfying
$f(1) = 2, f(2) = 3$ and $f'(x) \geq 1 \forall x \in (1,2)$.Define $g(x)=\int\limits_1^x {f(t)\,dt\,\forall \,x\, \in [1,2]} $ then the greatest value of $g(x)$ on $[1,2]$ is-
If the functions $f ( x )=\frac{ x ^3}{3}+2 bx +\frac{a x^2}{2}$ and $g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b$ have a common extreme point, then $a+2 b+7$ is equal to