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Period 2 from f(x + 1) = 1 + [2 - 3f + 3f^2 - f^3]^(1/3) | JEE Advanced

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

Question
If f(x)f(x) is a real valued function such that
f(x+1)=1+[2−3f(x)+3f2(x)−f3(x)]1/3,f\left(x+1\right) = 1+\left[2-3f(x)+3f^2(x)-f^3(x)\right]^{1/3} ,
 where f(x)≠0f(x) \ne 0 and f(x+2022)=2023λf(x)f\left(x+2022\right) = \dfrac{2023}{\lambda}f(x), then λ\lambda is equal to
Solution
Answer: 2023
Step 1:
1−3f+3f2−f3=(1−f)31-3f+3f^2-f^3 = \left(1-f\right)^3
 so
2−3f(x)+3f2(x)−f3(x)=1+(1−f(x))32-3f(x)+3f^2(x)-f^3(x) = 1+\left(1-f(x)\right)^3
 Step 2: Put h(x)=f(x)−1h(x) = f(x)-1, so 1−f(x)=−h(x)1-f(x) = -h(x) and the recurrence becomes
f(x+1)−1=[1−h3(x)]1/3⟹h(x+1)=[1−h3(x)]1/3f\left(x+1\right)-1 = \left[1-h^3(x)\right]^{1/3} \quad\Longrightarrow\quad h\left(x+1\right) = \left[1-h^3(x)\right]^{1/3}
h3(x+1)=1−h3(x)h^3\left(x+1\right) = 1-h^3(x)
 Step 3: Iterating once more,
h3(x+2)=1−h3(x+1)=1−(1−h3(x))=h3(x)h^3\left(x+2\right) = 1-h^3\left(x+1\right) = 1-\left(1-h^3(x)\right) = h^3(x)
 Cube roots are unique over R\mathbb R, so
h(x+2)=h(x)⟹f(x+2)=f(x)h\left(x+2\right) = h(x) \quad\Longrightarrow\quad f\left(x+2\right) = f(x)
 i.e. ff is periodic with period 22. Now 2022=2×10112022 = 2\times1011 is an even multiple of the period, so
f(x+2022)=f(x)f\left(x+2022\right) = f(x)
 Step 4: Comparing with the given relation,
f(x)=2023λf(x)for all xf(x) = \frac{2023}{\lambda}f(x) \qquad\text{for all } x
 and f(x)≠0f(x) \ne 0, so on cancelling f(x)f(x),
2023λ=1⟹λ=2023\frac{2023}\lambda = 1 \quad\Longrightarrow\quad \lambda = 2023
 Answer: 20232023
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