Application of DerivativeseasyFree

Non-Negative Integers Satisfying f(f(f(x))) > f(f(-x)) | JEE Advanced

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

Question
The number of non-negative integers which satisfy the inequation
f(f(f(x)))>f(f(−x)),f\left(f\left(f(x)\right)\right) > f\left(f(-x)\right),
 where f(x)=30−2x−x3f(x) = 30 - 2x - x^{3}, is
A00
B11
C22
D33correct
Solution
Step 1: Establish the key property of ff.
f′(x)=−2−3x2<0for every x,f'(x) = -2 - 3x^{2} < 0 \quad \text{for every } x ,
 so ff is strictly decreasing on R\mathbb{R}. Step 2: Record what that does to an inequality. For a strictly decreasing function,
f(A)>f(B)  ⟺  A<B,f(A) > f(B) \iff A < B ,
 i.e. applying ff reverses the inequality sign. Step 3: Strip off the outermost ff from both sides of f(f(f(x)))>f(f(−x))f\left(f(f(x))\right) > f\left(f(-x)\right):
f(f(x))<f(−x).f\left(f(x)\right) < f(-x) .
 Step 4: Strip off one more ff, reversing again.
f(x)>−x.f(x) > -x .
 Step 5: Substitute the formula for ff.
30−2x−x3>−x  ⟹  30−x−x3>0  ⟹  x3+x−30<0.30 - 2x - x^{3} > -x \implies 30 - x - x^{3} > 0 \implies x^{3} + x - 30 < 0 .
 Step 6: Factorise, spotting the root x=3x = 3.
x3+x−30=(x−3)(x2+3x+10)<0.x^{3}+x-30 = (x-3)\left(x^{2}+3x+10\right) < 0 .
 Step 7: Check the sign of the quadratic factor. Its discriminant is 9−40=−31<09 - 40 = -31 < 0 and its leading coefficient is positive, so x2+3x+10>0x^{2}+3x+10 > 0 always. The inequality therefore reduces to
x−3<0  ⟹  x<3.x - 3 < 0 \implies x < 3 .
 Step 8: Count the non-negative integers less than 33.
x=0, 1, 2  ⟹  3 values.x = 0,\ 1,\ 2 \implies 3 \ \text{values} .
 Answer: (4).
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 · 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
Application of Derivatives · hard
If xx and yy are real numbers such that x2+y2=27x^{2}+y^{2} = 27, then the maximum possible value of x−yx-y is

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