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Integers outside the domain of (f^2022 - h^2025)^(1/2024) | JEE Advanced

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

Question
f(x)f\left(x\right) is a continuous increasing function with f(x)≥xf\left(x\right) \ge x and h(x)=(1−x)−1h\left(x\right) = \left(1-x\right)^{-1}. Then the number of integral values in real numbers not belonging to the domain of
ϕ(x)=(f2022(x)−h2025(x))1/2024\phi\left(x\right) = \left(f^{2022}\left(x\right)-h^{2025}\left(x\right)\right)^{1/2024}
 is/are. (where f n(x)=f∘f∘⋯∘ff^{\,n}\left(x\right) = f\circ f\circ\cdots\circ f, nn times)
Solution
Answer: 2
Step 1:
h(x)=11−x,h2(x)=11−11−x=1−x−x=x−1xh\left(x\right) = \frac{1}{1-x},\qquad h^2\left(x\right) = \frac{1}{1-\frac{1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x}
h3(x)=11−x−1x=1x−(x−1)x=xh^3\left(x\right) = \frac{1}{1-\frac{x-1}{x}} = \frac{1}{\frac{x-\left(x-1\right)}{x}} = x
 So hh has order 33 under composition. Step 2:
2025=3×675⟹h2025(x)=x2025 = 3\times675 \quad\Longrightarrow\quad h^{2025}\left(x\right) = x
 Step 3: The chain breaks where any stage is undefined. hh needs x≠1x \ne 1; h2h^2 needs h(x)≠1h\left(x\right) \ne 1, i.e. 11−x≠1\dfrac{1}{1-x} \ne 1, i.e. x≠0x \ne 0; h3h^3 needs h2(x)≠1h^2\left(x\right) \ne 1, i.e. x−1x≠1\dfrac{x-1}{x} \ne 1, i.e. −1≠0-1 \ne 0, which never fails. As h3(x)=xh^3\left(x\right) = x, the next three stages repeat these same two conditions, and so on. Hence
domain of h2025=R∖{0,1}\text{domain of }h^{2025} = \mathbb{R}\setminus\left\{0,1\right\}
 Step 4: f(x)≥xf\left(x\right) \ge x and ff is increasing, so
f2(x)=f(f(x))≥f(x)≥xf^2\left(x\right) = f\left(f\left(x\right)\right) \ge f\left(x\right) \ge x
 and by induction f2022(x)≥xf^{2022}\left(x\right) \ge x for every xx. Hence
f2022(x)−h2025(x)=f2022(x)−x≥0f^{2022}\left(x\right)-h^{2025}\left(x\right) = f^{2022}\left(x\right)-x \ge 0
 The exponent 12024\tfrac1{2024} is an even root, so a non-negative radicand is exactly what is needed, and it is guaranteed everywhere. Step 5: The only obstruction is the domain of h2025h^{2025}, so
domain of ϕ=(−∞,0)∪(0,1)∪(1,∞)=R∖{0,1}\text{domain of }\phi = \left(-\infty,0\right)\cup\left(0,1\right)\cup\left(1,\infty\right) = \mathbb{R}\setminus\left\{0,1\right\}
{0,1}⟹2\left\{0,1\right\} \quad\Longrightarrow\quad 2
 Answer: 22
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