The range of the function f(x) = \(e^{x}+e^{x}\), is -
Step-by-Step Solution
Key Concept: Use \((\sec x + \tan x)(\sec x - \tan x) = 1\) . Put \(u = \sec x + \tan x\) .
<div class="solution"><p><strong>Key Idea:</strong> Use <span class="math-inline">\((\sec x + \tan x)(\sec x - \tan x) = 1\)</span>. Put <span class="math-inline">\(u = \sec x + \tan x\)</span>.</p><p><strong>Step 1:</strong> For <span class="math-inline">\(x\in(0,\pi/2)\)</span>, <span class="math-inline">\(u > 1\)</span>.</p><p><strong>Step 2:</strong> <span class="math-inline">\(\sec x - \tan x = 1/u\)</span>, so denominator <span class="math-inline">\(= 1 - 1/u = (u-1)/u\)</span></p><p><strong>Step 3:</strong> <span class="math-inline">\(f(x) = (u-1)/[(u-1)/u] = u = \sec x + \tan x\)</span></p><p><strong>Step 4:</strong> Range of <span class="math-inline">\(u\)</span> on <span class="math-inline">\((0,\pi/2)\)</span> is <span class="math-inline">\((1,\infty)\)</span></p><p><strong>Answer: <span class="math-inline">\((1,\infty)\)</span></strong></p><div class="trap-box"><strong>Trap:</strong> If you miss the reciprocal identity <span class="math-inline">\((\sec x+\tan x)(\sec x-\tan x)=1\)</span>, the simplification stays hidden.</div><div class="key-concept"><strong>Key Concept:</strong> sec-tan reciprocal identity for elegant simplification</div></div>
Correct Answer: C