Application of DerivativesmediumFree

Bound on f(4) from f(2) = -4 and f'(x) at least 6 | JEE Advanced LMVT

JEE Maths question with a full step-by-step solution.

Question
Let ff be a function which is continuous and differentiable for all real xx. If f(2)=−4f(2) = -4 and f′(x)≥6f'(x) \ge 6 for all x∈[2,4]x \in [2,4], then
Af(4)<8f(4) < 8
Bf(4)≥8f(4) \ge 8correct
Cf(4)≥12f(4) \ge 12
Dnone of these
Solution
Step 1: Check that Lagrange's mean value theorem applies. ff is continuous on [2,4][2,4] and differentiable on (2,4)(2,4), which is exactly what the theorem needs. Step 2: Write down what it gives. There exists c∈(2,4)c \in (2,4) with
f′(c)=f(4)−f(2)4−2.f'(c) = \frac{f(4)-f(2)}{4-2} .
 Step 3: Substitute the given value f(2)=−4f(2) = -4.
f′(c)=f(4)−(−4)2=f(4)+42.f'(c) = \frac{f(4)-(-4)}{2} = \frac{f(4)+4}{2} .
 Step 4: Apply the bound on the derivative. Since c∈(2,4)⊂[2,4]c \in (2,4) \subset [2,4], we have f′(c)≥6f'(c) \ge 6:
f(4)+42 ≥ 6.\frac{f(4)+4}{2} \ \ge\ 6 .
 Step 5: Solve for f(4)f(4).
f(4)+4≥12  ⟹  f(4)≥8.f(4)+4 \ge 12 \implies f(4) \ge 8 .
 Step 6: Check the options. (2) is exactly this. (1) contradicts it. (3) claims f(4)≥12f(4) \ge 12, which is stronger than what follows - for instance f(x)=6x−16f(x) = 6x - 16 has f(2)=−4f(2) = -4, f′(x)=6≥6f'(x) = 6 \ge 6 and f(4)=8<12f(4) = 8 < 12, so (3) can fail. Answer: (2).
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Application of Derivatives · medium
Let a function f:R→Rf : \mathbb{R} \to \mathbb{R} satisfy the functional equation f(x+y)=f(x)⋅f(y),f(x+y) = f(x)\cdot f(y), where f(0)≠0f(0) \ne 0 and f′(0)=1f'(0) = 1. Given g(x)=3xf(x),g(x) = \frac{3x}{f(x)}, then
Application of Derivatives · hard
Consider the function, for x∈[−2,3]x \in [-2, 3], f(x)={x3−2x2−5x+6x−1if x≠1−6if x=1f(x) = \begin{cases} \dfrac{x^{3}-2x^{2}-5x+6}{x-1} & \text{if } x \ne 1 \\[2mm] -6 & \text{if } x = 1 \end{cases} then
Application of Derivatives · hard
Let f(x)=lim⁡m→∞ln⁡(3+x2)−x2msin⁡(x2)1+x2mf(x) = \displaystyle\lim_{m\to\infty}\frac{\ln\left(3+x^2\right)-x^{2m}\sin\left(x^2\right)}{1+x^{2m}}, then which of the following is/are CORRECT?
Application of Derivatives · medium
For the functions f(θ)=αtan⁡2θ+βcot⁡2θf(\theta)=\alpha\tan^2\theta+\beta\cot^2\theta and g(θ)=αsin⁡2θ+βcos⁡2θg(\theta)=\alpha\sin^2\theta+\beta\cos^2\theta, α>β>0\alpha>\beta>0, let min⁡0<θ<π/2f(θ)=max⁡0<θ<πg(θ)\displaystyle\min_{0<\theta<\pi/2}f(\theta)=\max_{0<\theta<\pi}g(\theta). If the first term of a G.P. is (α2β)\left(\dfrac{\alpha}{2\beta}\right), its common ratio is (2βα)\left(\dfrac{2\beta}{\alpha}\right), and the sum of its first 1010 terms is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath