Q.If the function $f(x) = ax^3 + bx^2 + 11x - 6$ satisfies conditions of Rolle's theorem in $[1, 3]$ and $f'\left(2 + \dfrac{1}{\sqrt3}\right) = 0$, then values of $a$ and $b$ are respectively. (A) $1, -6$ (B) $-2, 1$ (C) $-1, -6$ (D) $-1, 6$

Rolle's Theorem states that if a real-valued function $f$ is continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a)=f(b)$, then there exists at least one point $c$ in $(a,b)$ where $f'(c)=0$. Geometrically, if a smooth curve starts and ends at the same height, somewhere in between it must have a point where the tangent is horizontal (parallel to the X-axis) — the curve must turn around. To apply the theorem to a given function on a given interval: first check the three hypotheses (the function must be continuous on the closed interval and differentiable on the open interval — true automatically for polynomials, and for combinations of standard continuous/differentiable functions such as $e^x$, $\sin x$, $\cos x$, $\log x$ wherever they are individually defined and smooth; and $f(a)$ must equal $f(b)$, checked by direct substitution). If any hypothesis fails (as with a function that is not differentiable at an interior point, or one where the two endpoint values differ), Rolle's Theorem does not apply. If all hypotheses hold, differentiate $f$, set $f'(x)=0$, solve for $x$, and confirm the solution(s) $c$ actually lie strictly inside $(a,b)$.

"Rolle's theorem statement and verification examples" and "Rolle's theorem class 12 important questions" are frequently searched around the Continuity and Differentiability chapter of the NCERT-aligned CBSE Class 12 Mathematics curriculum, a topic tested in board exams and JEE Main mean-value-theorem questions. Checking all three hypotheses carefully before applying the theorem, as emphasised here, is exactly what separates full marks from a partial answer in competitive-exam solutions.