Basics & LogarithmshardFree

If (1 + 1 by x) to the power (x + 1) = (1 + 1 by 2021) to the power 2021, find mod x | JEE Main

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Question
If (1+1x)x+1=(1+12021)2021\left(1+\dfrac1x\right)^{x+1} = \left(1+\dfrac{1}{2021}\right)^{2021}, then ∣x∣\left|x\right| is equal to
Solution
Answer: 2022
Step 1: (1+1x)x+1\left(1 + \dfrac{1}{x}\right)^{x+1} is defined for 1+1x>01 + \dfrac{1}{x} > 0, i.e. x>0x > 0 or x<−1x < -1. Step 2:
(1+12021)2021=(20222021)2021=(20212022)−2021\left(1 + \frac{1}{2021}\right)^{2021} = \left(\frac{2022}{2021}\right)^{2021} = \left(\frac{2021}{2022}\right)^{-2021}
 Step 3: 20212022=1−12022=1+1−2022\dfrac{2021}{2022} = 1 - \dfrac{1}{2022} = 1 + \dfrac{1}{-2022} and −2021=(−2022)+1-2021 = (-2022) + 1, so
(1+12021)2021=(1+1−2022)(−2022)+1\left(1 + \frac{1}{2021}\right)^{2021} = \left(1 + \frac{1}{-2022}\right)^{(-2022)+1}
 ∴ x=−2022\therefore\ x = -2022 satisfies the given equation, and −2022<−1-2022 < -1. Step 4: Let f(x)=(x+1)ln⁡(1+1x)f(x) = (x+1)\ln\left(1 + \dfrac{1}{x}\right).
f′(x)=ln⁡(1+1x)+(x+1)(1x+1−1x)=ln⁡(1+1x)−1xf'(x) = \ln\left(1 + \frac{1}{x}\right) + (x+1)\left(\frac{1}{x+1} - \frac{1}{x}\right) = \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x}
 Putting u=1xu = \dfrac{1}{x}, we have ln⁡(1+u)<u\ln(1+u) < u for u>−1, u≠0u > -1,\ u \ne 0, so f′(x)<0f'(x) < 0 on x>0x > 0 and on x<−1x < -1. Step 5: For x>0x > 0, ff is decreasing and lim⁡x→∞f(x)=1\displaystyle\lim_{x \to \infty} f(x) = 1, so f(x)>1f(x) > 1, i.e.
(1+1x)x+1>efor x>0\left(1 + \frac{1}{x}\right)^{x+1} > e \quad \text{for } x > 0
 But (1+12021)2021<e\left(1 + \dfrac{1}{2021}\right)^{2021} < e, which is not possible. Hence there is no root with x>0x > 0. Step 6: On x<−1x < -1, ff is strictly decreasing, so the root there is unique. Therefore x=−2022x = -2022 is the only root and
∣x∣=∣−2022∣=2022|x| = |-2022| = 2022
 Answer: 20222022.
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