Limits, Continuity & Differentiability
Continuity and function properties
Grade 12
Question:
<p>If <i>f</i>(<i>x</i>) is continuous such that |<i>f</i>(<i>x</i>)| ≤ 1, ∀<i>x</i> ∈ ℝ and \(\frac{e^{f(x)} - e^{|f(x)|}}{e^{f(x)} + e^{|f(x)|}} ≥ 0\), then range of <i>g</i>(<i>x</i>) is \(g(x) = \frac{e^{f(x)} - e^{-|f(x)|}}{e^{f(x)} + e^{-|f(x)|}}\)</p>
<p>(a) \([0, 1]\)</p>
<p>(b) \(\left[0, \frac{e^2-1}{e^2+1}\right]\)</p>
<p>(c) \(\left[0, \frac{e^2-1}{e^2+1}\right]\)</p>
<p>(d) \(\left[\frac{1-e}{1+e}, 0\right]\)</p>
Step-by-Step Solution
Key Concept: Analyze the inequality constraint to determine the valid range of f(x), then evaluate g(x) at boundary values.
<p>Given |<i>f</i>(<i>x</i>)| ≤ 1 and $\frac{e^{f(x)} - e^{|f(x)|}}{e^{f(x)} + e^{|f(x)|}} ≥ 0$. This implies $e^{f(x)} ≥ e^{|f(x)|}$, which means <i>f</i>(<i>x</i>) ≥ |<i>f</i>(<i>x</i>)|. This is only possible when <i>f</i>(<i>x</i>) ≥ 0. Combined with |<i>f</i>(<i>x</i>)| ≤ 1, we have 0 ≤ <i>f</i>(<i>x</i>) ≤ 1. The function $g(x) = \frac{e^{f(x)} - e^{-|f(x)|}}{e^{f(x)} + e^{-|f(x)|}}$ is increasing in <i>f</i>(<i>x</i>), so its range is $[g(0), g(1)] = \left[0, \frac{e^2-1}{e^2+1}\right]$.</p>
Correct Answer: B