Free JEE Main Maths PYQ Rolle's and Lagrange's Theorem MCQs with step-by-step solutions (2 questions). Part of Application of Derivatives. Practise online on Prepizo — no login needed.
Q1 — Rolle's and Lagrange's Theorem · hard · theory
If $f:R\to R$ is a twice differentiable function such that $f''(x)\ge0$ for all $x\in R$, and $f\left(\dfrac12\right)=\dfrac12$, $f(1)=1$, then
A. $f'(1)\le0$
B. $f'(1)>1$ ✓ Correct
C. $0<f'(1)\le\dfrac12$
D. $\dfrac12<f'(1)\le1$
Solution: Since $f''\ge0$, $f'$ is non-decreasing (convex $f$). By the Mean Value Theorem on $\left[\tfrac12,1\right]$, there is $c\in\left(\tfrac12,1\right)$ with $f'(c)=\dfrac{f(1)-f(1/2)}{1-1/2}=1$. As $f'$ is non-decreasing, $f'(1)\ge f'(c)=1$; since $c<1$ and equality throughout would force $f$ linear (contradicting the strict convexity needed to pin $c$ inside the open interval), the inequality is strict: $f'(1)>1$.
Q2 — Rolle's and Lagrange's Theorem · medium · theory
Let $f(x)=2-\cos x$, for all real $x$. Statement I For each real $t$, there exists a point $c$ in $[t,t+2\pi]$, such that $f'(c)=0$. Statement II $f(t)=f(t+2\pi)$ for each real $t$.
A. Statement I is correct, Statement II is also correct; Statement II is the correct explanation of Statement I
B. Statement I is correct, Statement II is also correct; Statement II is not the correct explanation of Statement I ✓ Correct
C. Statement I is correct; Statement II is incorrect
D. Statement I is incorrect; Statement II is correct
Solution: Since $\cos x$ has period $2\pi$, $f(t)=f(t+2\pi)$ for every real $t$, so Statement II is true. As $f$ is differentiable and $f(t)=f(t+2\pi)$, Rolle's theorem then guarantees a point $c\in(t,t+2\pi)$ with $f'(c)=0$, so Statement I is also true. But Statement II by itself is only the hypothesis used inside Rolle's theorem, not the full reasoning, so it is not accepted as the complete explanation of Statement I.