Q.Let $f(x)$ and $g(x)$ be differentiable for $0 < x < 1$ such $f(0) = 0, g(0) = 0, f(1) = 6$. Let there exist a real number $c$ in $(0, 1)$ such that $f'(c) = 2g'(c)$, then the value of $g(1)$ must be (A) 1 (B) 3 (C) 2.5 (D) $-1$

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 …