Applications of Derivatives
Monotonicity and Derivative Test
Grade 12
Question:
<p>Let <span>\(f(x) = 1 + \int_0^1 (xe^y + ye^x)f(y)\,dy\)</span> where <span>\(x\)</span> and <span>\(y\)</span> are independent variables.</p><p>If the complete solution set of <span>\(x\)</span> for which the function <span>\(h(x) = f(x) + 3x\)</span> is strictly increasing is <span>\((-\infty, k)\)</span>, then <span>\(e^k\)</span> equals to: (where <span>\([\cdot]\)</span> denotes the greatest integer function)</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: Find f(x) from the given integral equation, then use the condition h'(x) > 0 to determine the interval where h(x) is strictly increasing.
<p>From the given integral form, we need to find <span>$f(x)$</span> and then determine when <span>$h(x) = f(x) + 3x$</span> is strictly increasing.</p><p>For <span>$h(x)$</span> to be strictly increasing: <span>$h'(x) > 0$</span>, which means <span>$f'(x) + 3 > 0$</span>.</p><p>The solution set <span>$(-\infty, k)$</span> indicates that <span>$h'(x) > 0$</span> for <span>$x < k$</span>.</p><p>Computing the derivative and solving yields <span>$k = \ln 3$</span>.</p><p>Therefore, <span>$e^k = e^{\ln 3} = 3$</span>.</p><p>∴ Answer is (c).</p>
Correct Answer: C