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9709 P11 - Jun 2019 - Q11
1274
The diagram shows part of the curve \(y = \frac{3}{\sqrt{1 + 4x}}\) and a point \(P(2, 1)\) lying on the curve. The normal to the curve at \(P\) intersects the \(x\)-axis at \(Q\).
(i) Show that the \(x\)-coordinate of \(Q\) is \(\frac{16}{9}\).
(ii) Find, showing all necessary working, the area of the shaded region.
Solution
(i) To find the \(x\)-coordinate of \(Q\), we first find the derivative of the curve \(y = \frac{3}{\sqrt{1 + 4x}}\). The derivative is \(\frac{dy}{dx} = 3 \times -\frac{1}{2} \times (1 + 4x)^{-\frac{3}{2}} \times 4\).
At \(x = 2\), the gradient \(m\) is \(-\frac{2}{9}\). The perpendicular gradient (normal) is \(\frac{9}{2}\).
The equation of the normal is \(y - 1 = \frac{9}{2}(x - 2)\).
Setting \(y = 0\) to find \(Q\), we solve \(0 - 1 = \frac{9}{2}(x - 2)\), giving \(x = \frac{16}{9}\).
(ii) The area under the curve from \(x = 0\) to \(x = 2\) is \(\int_{0}^{2} \frac{3}{\sqrt{1 + 4x}} \, dx = \frac{3}{2} \sqrt{1 + 4x} \bigg|_0^2 \times \frac{1}{4}\).
Evaluating gives \(3\).
The area of the triangle is \(\frac{1}{2} \times 1 \times \frac{2}{9} = \frac{1}{9}\).
The shaded area is \(3 - \frac{1}{9} = \frac{8}{9}\).
