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9709 P13 - Jun 2013 - Q11
1295
The diagram shows part of the curve \(y = \frac{8}{\sqrt{x}} - x\) and points \(A (1, 7)\) and \(B (4, 0)\) which lie on the curve. The tangent to the curve at \(B\) intersects the line \(x = 1\) at the point \(C\).
(i) Find the coordinates of \(C\).
(ii) Find the area of the shaded region.
Solution
(i) To find the coordinates of \(C\), we first find the derivative of the curve \(y = \frac{8}{\sqrt{x}} - x\). The derivative is \(\frac{dy}{dx} = -4x^{-\frac{3}{2}} - 1\). Substituting \(x = 4\) into the derivative gives \(\frac{dy}{dx} = -\frac{3}{2}\). The equation of the tangent at \(B(4, 0)\) is \(y - 0 = -\frac{3}{2}(x - 4)\). Solving for \(y\) when \(x = 1\), we find \(C(1, \frac{4}{2})\).
(ii) To find the area of the shaded region, we calculate the area under the curve from \(x = 1\) to \(x = 4\). The integral is \(\int \left( \frac{8}{\sqrt{x}} - x \right) \, dx\), which evaluates to \(8x^{\frac{1}{2}} - \frac{1}{2}x^2\). Using limits 1 to 4, the area under the curve is \(8\frac{1}{2}\). The area under the tangent is \(\frac{1}{2} \times \frac{4}{2} \times 3 = 6\frac{3}{4}\). The shaded area is \(\frac{13}{4}\).
