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2018 Paper 1 Q1
D: 1516.0 B: 1516.0
polynomials curve sketching tangent-normal integration area between curves inequality quadratic/cubic intersections triangle area

The line \(y=a^2 x\) and the curve \(y=x(b-x)^2\), where \(0 < a < b\,\), intersect at the origin \(O\) and at points \(P\) and \(Q \). The \(x\)-coordinate of \(P\) is less than the \(x\)-coordinate of \(Q\). Find the coordinates of \(P\) and \(Q\), and sketch the line and the curve on the same axes. Show that the equation of the tangent to the curve at \(P\) is \[ y = a(3a-2b)x + 2a(b-a)^2 . \] This tangent meets the \(y\)-axis at \(R\). The area of the region between the curve and the line segment \(OP\) is denoted by \(S\). Show that \[ S= \frac1{12}(b-a)^3(3a+b)\,. \] The area of triangle \(OPR\) is denoted by \(T\). Show that \(S>\frac{1}{3}T\,\).

Solution:

TikZ diagram
\begin{align*} && a^2x &= x(b-x)^2 \\ \Rightarrow && 0 &= x((b-x)^2-a^2) \\ &&&= x(b-a-x)(b+a-x)\\ && y &= x(b-x)^2 \\ \Rightarrow && y' &= (b-x)^2-2x(b-x) \\ P(b-a,a^2(b-a)): &&y' &= (b-(b-a))^2-2(b-a)(b-(b-a)) \\ &&&= a^2-2a(b-a) = a(3a-2b) \\ \Rightarrow && y &= a(3a-2b)(x-(b-a)) + a^2(b-a) \\ &&&= a(3a-2b)x + (b-a)(a^2-3a^2+2ba) \\ &&&= a(3a-2b)x + (b-a)2a(b-a) \\ &&&= a(3a-2b)x + 2a(b-a)^2 \\ \end{align*} Therefore the tangent at \(P\) is \(a(3a-2b)x + 2a(b-a)^2\) The area between the curve and \(OP\) is \begin{align*} &&S &= \int_0^{b-a} \left (x(b-x)^2-a^2x \right) \d x\\ &&&= \left [\frac{x^2}{2}b^2 - \frac{2x^3}{3}b +\frac{x^4}{4} - \frac{a^2x^2}{2}\right]_0^{b-a} \\ &&&= (b-a)^2 \tfrac12 (b^2-a^2) - \tfrac23(b-a)^3b + \tfrac14(b-a)^4 \\ &&&= \tfrac1{12}(b-a)^3(6(b+a)-8b+3(b-a)) \\ &&&= \tfrac1{12}(b-a)^3(b+3a) \end{align*} The area \([OPR] = T= \tfrac12 \cdot (b-a) \cdot 2a(b-a)^2 = a(b-a)^3\) Clearly \(S > \frac4{12}(b-a)^3a = \frac13T\)

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