June 2023 Paper 2 Q4
4 A curve has equation
\[y = \frac{x^2}{8} + 4\sqrt{x}\](a) Find an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) [3 marks]
(b) The point \(P\) with coordinates \((4, 10)\) lies on the curve.
Find an equation of the tangent to the curve at the point \(P\) [2 marks]
(c) Show that the curve has no stationary points. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes \(\sqrt{x}\) as \(x^{\frac{1}{2}}\) PI by derivative with \(kx^{-\frac{1}{2}}\) | B1 | 1.1b |
| Differentiates with at least one term correct | M1 | 1.1a |
| Obtains a correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) ACF ISW | A1 | 1.1b |
| (3) |
Typical solution
\[y = \frac{x^2}{8} + 4x^{\frac{1}{2}}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{4} + 2x^{-\frac{1}{2}}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains gradient of 2 or Substitutes \(x\) = 4 into their expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 | 1.1a |
| Obtains correct equation of the tangent. Does not need to be fully simplified. ACF For example: \(y = 2x + 2\) ISW | A1 | 1.1b |
| (2) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 2\]\[y - 10 = 2(x - 4)\]| Scheme | Marks | AO |
|---|---|---|
| Equates their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to zero. OE | M1 | 1.1a |
| Completes reasoned argument by correctly manipulating the equation to obtain or \(x^{\frac{3}{2}} = -8\) or \(x^{\frac{1}{2}} = -2\) and states \(x = 4\) is a solution and then deduces \(\dfrac{x}{4} + \dfrac{2}{\sqrt{x}} = 2 \neq 0\) or the gradient found at \(x = 4\) in part (b) was non-zero and concludes that the curve has no stationary points. or Completes reasoned argument by correctly manipulating the equation to obtain \(x^2 = -8\sqrt{x}\) or \(x^{\frac{3}{2}} = -8\) or \(x^{\frac{1}{2}} = -2\) and deduces the equation has no solutions by making explicit reference to \(\sqrt{x} \gt 0\) and concludes that the curve has no stationary points. or Completes reasoned argument to establish that \(\dfrac{x}{4} + \dfrac{2}{\sqrt{x}} \gt 0\) and deduces the equation has no solutions and concludes that the curve has no stationary points. | R1 | 2.1 |
| (2) | ||
| (7 marks) |
Typical solution
\[\frac{x}{4} + 2x^{-\frac{1}{2}} = 0\]\[\frac{x}{4} + \frac{2}{\sqrt{x}} = 0\]As \(x \gt 0\),
\[\frac{2}{\sqrt{x}} \gt 0 \text{ and } \frac{x}{4} \gt 0\]\[\text{Therefore } \frac{x}{4} + \frac{2}{\sqrt{x}} \gt 0\]Equation has no solutions so the curve has no stationary points.