AS June 2025 Paper 1 Q13
13 The curve \(C_1\) is given by the equation
\[xy = m\]where \(m\) is a positive constant.
The region \(R_1\) is enclosed by curve \(C_1\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\)
The region \(R_1\) is rotated through \(2\pi\) radians about the \(x\)-axis.
The volume of the solid generated is \(V\)
(a) Find a simplified expression for \(V\) in terms of \(m\) [2 marks]
(b) The curve \(C_2\) is a stretch of \(C_1\) by a factor 5 in the \(y\)-direction.
The region \(R_2\) is enclosed by \(C_2\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\)
The region \(R_2\) is rotated through \(2\pi\) radians about the \(x\)-axis.
The volume of the solid generated by rotating \(R_2\) is \(6\pi\)
Calculate the value of \(m\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes a correct expression for \(V\) in terms of \(x\) Condone missing \(\mathrm{d}x\) | M1 | 1.2 |
| Obtains \(\dfrac{3}{4}m^2\pi\) | A1 | 1.1b |
| (2) |
Typical solution
\[\begin{aligned} V &= \pi\int_1^4 \left(\frac{m}{x}\right)^2 \mathrm{d}x \\ &= \pi m^2\left[-\frac{1}{x}\right]_1^4 \\ &= \frac{3\pi m^2}{4}\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Writes a correct expression for the stretched volume in terms of \(m\) (and \(x\)) eg \(\pi\displaystyle\int_1^4 \left(\frac{5m}{x}\right)^2 \mathrm{d}x\) or Writes their part (a) \(\times\, 5^2\) | M1 | 3.1a |
| Forms a correct equation in terms of \(m\) only or Forms the equation their part (a) \(\times\, 5^n = 6\pi\) where \(n = 1, 2\) or 3 PI by \(m^2 = \dfrac{8}{5}\) or \(m^2 = \dfrac{8}{125}\) oe | M1 | 1.1a |
| Obtains \(\dfrac{2}{5}\sqrt{2}\) oe eg \(\dfrac{\sqrt{8}}{5}\) Negative root does not have to be considered. | A1 | 1.1b |
| (3) | ||
| (5 marks) |