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Real solutions of (x^2024 + 1)(1 + x^2 + ...) = 2024 x^2023 | JEE Advanced

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

Question
The number of real solutions of the equation
(x2024+1)(1+x2+x4+⋯+x2022)=2024 x2023\left(x^{2024}+1\right)\left(1+x^2+x^4+\cdots+x^{2022}\right) = 2024\,x^{2023}
 is
Solution
Answer: 1
Step 1: Putting x=0x = 0 gives 1⋅1=01 \cdot 1 = 0, which is not possible, so x≠0x \ne 0 and we may divide by x2023x^{2023}.
x2024+1x2023(1+x2+⋯+x2022)=2024\frac{x^{2024}+1}{x^{2023}}\left(1+x^2+\cdots+x^{2022}\right) = 2024
(x+1x2023)(1+x2+x4+⋯+x2022)=2024\left(x+\frac1{x^{2023}}\right)\left(1+x^2+x^4+\cdots+x^{2022}\right) = 2024
 Step 2:
x(1+x2+⋯+x2022)=x+x3+x5+⋯+x2023x\left(1+x^2+\cdots+x^{2022}\right) = x+x^3+x^5+\cdots+x^{2023}
1x2023(1+x2+⋯+x2022)=1x2023+1x2021+⋯+1x\frac1{x^{2023}}\left(1+x^2+\cdots+x^{2022}\right) = \frac1{x^{2023}}+\frac1{x^{2021}}+\cdots+\frac1{x}
 so the equation is
(x+1x)+(x3+1x3)+⋯+(x2023+1x2023)=2024\left(x+\frac1x\right)+\left(x^3+\frac1{x^3}\right)+\cdots+\left(x^{2023}+\frac1{x^{2023}}\right) = 2024
 There are 10121012 brackets, one for each odd exponent 1,3,…,20231,3,\dots,2023. Step 3: For x<0x<0 every odd power x2k−1x^{2k-1} is negative and so is its reciprocal, so the left side is negative and never 20242024. Hence x>0x>0. Step 4: By A.M.-G.M., for each odd mm and each x>0x>0,
xm+1xm ≥ 2x^m+\frac1{x^m} \ \ge\ 2
 with equality if and only if xm=1x^m = 1, i.e. x=1x = 1. Adding the 10121012 brackets,
LHS ≥ 2×1012=2024\text{LHS} \ \ge\ 2\times1012 = 2024
 with equality only at x=1x = 1. Step 5:
(1+1)(1+1+⋯+1⏟1012)=2×1012=2024.\left(1+1\right)\left(\underbrace{1+1+\cdots+1}_{1012}\right) = 2\times1012 = 2024 .
 So there is exactly one real solution. Answer: 11
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