The range of the function f : \(\mathbb{N} \to \mathbb{Z}\); f(x) = \((1-x)^{x-1}\), is -
Step-by-Step Solution
Key Concept: Let $a=\sqrt{x+2},\ b=\sqrt{2-x}$ . Then $a^2+b^2=4$ .
<div class="solution"><p><strong>Key Idea:</strong> Let <span class="math-inline">$a=\sqrt{x+2},\ b=\sqrt{2-x}$</span>. Then <span class="math-inline">$a^2+b^2=4$</span>.</p><p><strong>Step 1:</strong> Domain: <span class="math-inline">$x\in[-2,2]$</span></p><p><strong>Step 2:</strong> Since <span class="math-inline">$a^2+b^2=4$</span>, the sum <span class="math-inline">$a+b$</span> satisfies <span class="math-block">$$\sqrt{2} \le a+b \le 2\sqrt{2}$$</span></p><p><strong>Step 3:</strong> <span class="math-block">$$\frac{4}{a+b} \in [\sqrt{2},\ 2]$$</span></p><p><strong>Step 4:</strong> <span class="math-inline">$f(x) = \log_2$</span> of that, giving range <span class="math-inline">$[\frac{1}{2}, 1]$</span></p><p><strong>Answer: <span class="math-inline">$\left[\frac{1}{2},1\right]$</span></strong></p><div class="trap-box"><strong>Trap:</strong> Do not differentiate immediately. The fixed square-sum <span class="math-inline">$a^2+b^2=4$</span> makes a range argument much cleaner.</div><div class="key-concept"><strong>Key Concept:</strong> Range via substitution + AM-QM on constrained variables</div></div>
Correct Answer: B