Applications of Derivatives Question - None

Question

Paragraph for Questions 595 and 596:

A quadratic polynomial $ f(x) $ with positive leading coefficient such that $ g(x) = f(\ln x) \ \forall x > 0 $. Also the curve $ y = g(x) $ satisfies the following conditions:

(a) There is exactly one value for a positive number $ p $ such that $ (p, g(p)) $ is its extremum point and $ (p^2, g(p^2)) $ is its inflection point.

(b) Exactly one tangent line can be drawn from the point $ (0, 0) $ to the curve $ y = g(x) $.

The value of $ \dfrac{f(10)}{f(2)} $ equals:

Accepted Answer

Step 1: Let f(x) = ax² + bx + c with a > 0. Then g(x) = f(ln x) = a(ln x)² + b(ln x) + c.

Step 2: Compute derivatives: g'(x) = (2a ln x + b)/x and g''(x) = (2a - 2a ln x - b)/x².

Step 3: At extremum point p: g'(p) = 0 ⟹ 2a ln p + b = 0 ⟹ b = -2a ln p.

Step 4: At inflection point p²: g''(p²) = 0 ⟹ 2a - 2a ln(p²) - b = 0 ⟹ 2a - 4a ln p - b = 0.

Step 5: Substitute b = -2a ln p into the inflection condition: 2a - 4a ln p + 2a ln p = 0 ⟹ 2a(1 - ln p) = 0 ⟹ ln p = 1 ⟹ p = e.

Step 6: Thus b = -2a and g(x) = a(ln x)² - 2a ln x + c = a(ln x - 1)² + (c - a).

Step 7: For exactly one tangent from origin to y = g(x): If tangent at point (t, g(t)) passes through (0,0), then g(t)/t = g'(t). This gives: [a(ln t - 1)² + (c - a)]/t = (2a ln t - 2a)/t.

Step 8: Simplifying: a(ln t - 1)² + (c - a) = 2a(ln t - 1) ⟹ a(ln t - 1)² - 2a(ln t - 1) + (c - a) = 0.

Step 9: Let u = ln t - 1: au² - 2au + (c - a) = 0. For exactly one tangent, discriminant = 0: 4a² - 4a(c - a) = 0 ⟹ 4a² - 4ac + 4a² = 0 ⟹ c = 2a.

Step 10: Therefore f(x) = a(x² - 2x + 2). Verify: g(x) = a(ln x - 1)² + a with minimum at x = e and inflection at x = e².

Step 11: Calculate the ratio: f(10)/f(2) = [a(100 - 20 + 2)]/[a(4 - 4 + 2)] = 82/2 = 41.

∴ Answer: A

Question:

Step-by-Step Solution

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