Suppose that $f (0) = - 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is
Suppose that $f (0) = - 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is
- A
$7$
- B
$- 7$
- C
$13$
- D
$8$
Similar Questions
The function $f(x) = {(x - 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x - 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$ and $(4, 1)$ is
The function $f(x) = {(x - 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x - 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$ and $(4, 1)$ is
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 ?
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 for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is
If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is
lf Rolle's theorem holds for the function $f(x) =2x^3 + bx^2 + cx, x \in [-1, 1],$ at the point $x = \frac {1}{2},$ then $2b+ c$ equals
lf Rolle's theorem holds for the function $f(x) =2x^3 + bx^2 + cx, x \in [-1, 1],$ at the point $x = \frac {1}{2},$ then $2b+ c$ equals
- [JEE MAIN 2015]
If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............
If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............