Question:
The minimum of $f(x) = \dfrac{x^{100} - 1}{x^{100} + 1}, \; x \in \mathbb{R}$ is:
The minimum of $f(x) = \dfrac{x^{100} - 1}{x^{100} + 1}, \; x \in \mathbb{R}$ is:
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Replace even powers with variable to simplify analysis.
Updated On: Apr 24, 2026
- $-5$
- $-1.5$
- $-1$
- $-2$
- $-3$
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The Correct Option is C
Solution and Explanation
Concept:
• Even powers: $x^{100} \geq 0$
Step 1: Let $t = x^{100}$
\[ t \geq 0 \] \[ f(x) = \frac{t-1}{t+1} \]
Step 2: Analyze function
\[ f(t) = \frac{t-1}{t+1} \] As $t \to 0$: \[ f = \frac{-1}{1} = -1 \] As $t \to \infty$: \[ f \to 1 \]
Step 3: Conclusion
Minimum value occurs at $t=0$. Final Conclusion:
\[ = -1 \]
• Even powers: $x^{100} \geq 0$
Step 1: Let $t = x^{100}$
\[ t \geq 0 \] \[ f(x) = \frac{t-1}{t+1} \]
Step 2: Analyze function
\[ f(t) = \frac{t-1}{t+1} \] As $t \to 0$: \[ f = \frac{-1}{1} = -1 \] As $t \to \infty$: \[ f \to 1 \]
Step 3: Conclusion
Minimum value occurs at $t=0$. Final Conclusion:
\[ = -1 \]
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