If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$,then $a+b=$

  • A
    -$5$
  • B
    -$3$
  • C
    $4$
  • D
    $7$

Explore More

Similar Questions

Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

Values of $c$ as per Rolle's theorem for $f(x)=\sin x+\cos x+6$ on $[0, 2\pi]$ are

If $a, b, c$ are real numbers such that $a + b + c = 0$,then the quadratic equation $3ax^2 + 2bx + c = 0$ has

Difficult
View Solution

Given that $f(x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f(a) < 0$ and $f(b) > 0$,which of the following are always true?
$(i)$ $f(x)$ is bounded on $a \le x \le b$.
$(ii)$ The equation $f(x) = 0$ has at least one solution in $a < x < b$.
$(iii)$ The maximum and minimum values of $f(x)$ on $a \le x \le b$ occur at points where $f'(c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f'(c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f'(d) < 0$.

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo