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9709 P11 - Jun 2023 - Q11
1256
The equation of a curve is such that \(\frac{dy}{dx} = 6x^2 - 30x + 6a\), where \(a\) is a positive constant. The curve has a stationary point at \((a, -15)\).
(a) Find the value of \(a\).
(b) Determine the nature of this stationary point.
(c) Find the equation of the curve.
(d) Find the coordinates of any other stationary points on the curve.
Solution
(a) To find the value of \(a\), set \(\frac{dy}{dx} = 0\) at the stationary point \(x = a\). Thus, \(6a^2 - 30a + 6a = 0\). Simplifying gives \(6a(a - 4) = 0\), so \(a = 4\).
(b) Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2} = 12x - 30\). At \(x = 4\), \(\frac{d^2y}{dx^2} = 18\), which is positive, indicating a minimum.
(c) Integrate \(\frac{dy}{dx} = 6x^2 - 30x + 24\) to find \(y\). The integral is \(y = 2x^3 - 15x^2 + 24x + c\). Using the point \((4, -15)\), solve for \(c\): \(-15 = 2(4)^3 - 15(4)^2 + 24(4) + c\), giving \(c = 1\). Thus, \(y = 2x^3 - 15x^2 + 24x + 1\).
(d) Set \(\frac{dy}{dx} = 0\) to find other stationary points: \(6x^2 - 30x + 24 = 0\). Factor to get \(6(x-1)(x-4) = 0\), so \(x = 1\) or \(x = 4\). For \(x = 1\), \(y = 2(1)^3 - 15(1)^2 + 24(1) + 1 = 12\). Thus, the coordinates are \((1, 12)\).
