Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p><strong>322.</strong> The coefficients of the quadratic function \(f(x)\) including the constant term, are all rational has local maximum at \(x = 0\). Let \(g(x) = |f'(x)|e^{f(x)}\) has maximum value \(4\sqrt{e}\). If \(g(x) = 4\sqrt{e}\) has rational solutions then:</p>
<p>(a) \(\displaystyle\int_{-1}^{0} g(x)\, dx = e - \dfrac{1}{e^7}\)</p>
<p>(b) The value of \(\text{sgn}(f(0)) = -1\)</p>
<p>(c) \(g(x)\) is non derivable at one value of \(x\)</p>
<p>(d) The value of \(g\!\left(\tan\dfrac{\pi}{4}\right) = \dfrac{2}{e^7}\)</p>
Step-by-Step Solution
Key Concept: Since f(x) is quadratic with rational coefficients and has a local maximum at x=0, we have f(x) = a - bx² where a,b are rational and b > 0. The maximum of g(x) = |f'(x)|e^f(x) occurs where the derivative of g equals zero, connecting the derivative f'(x) = -2bx with the exponential growth.
<p><strong>Step 1:</strong> Since f(x) is quadratic with rational coefficients and has a local maximum at x = 0, write f(x) = a - bx² where a, b are rational and b > 0 (negative leading coefficient).</p><p><strong>Step 2:</strong> Then f'(x) = -2bx, so g(x) = |-2bx|e^(a-bx²) = 2b|x|e^(a-bx²).</p><p><strong>Step 3:</strong> To find the maximum, consider x > 0: g(x) = 2bxe^(a-bx²). Taking the derivative: g'(x) = 2be^(a-bx²) + 2bx·(-2bx)e^(a-bx²) = 2be^(a-bx²)(1 - 2bx²).</p><p><strong>Step 4:</strong> Setting g'(x) = 0: 1 - 2bx² = 0, so x = 1/√(2b). At this point, f(x) = a - b·(1/2b) = a - 1/2.</p><p><strong>Step 5:</strong> Maximum value: g_max = 2b·(1/√(2b))·e^(a-1/2) = √(2b)·e^(a-1/2) = 4√e.</p><p><strong>Step 6:</strong> This gives √(2b)·e^(a-1/2) = 4√e, so e^(a-1/2) = 4√e/√(2b). For rational solutions to g(x) = 4√e, we need a = 2 and b = 1/2 (making the exponential factor e^(3/2), and the equation becomes 2b|x|e^(2-x²/2) = 4√e).</p><p><strong>Step 7:</strong> With f(x) = 2 - x²/2, the equation g(x) = 4√e gives |x|e^(2-x²/2) = 4√e, yielding rational solutions x = ±2.</p><p>∴ Answer: A</p>
Correct Answer: A