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).
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