| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Line tangent to curve, find k for tangency |
| Difficulty | Standard +0.3 This is a slightly above-average P1 question. Part (i) requires solving a quadratic equation after substituting and squaring. Part (ii) requires using the discriminant condition (b²-4ac=0) for tangency, which is a standard technique but requires students to recognize when to apply it. The algebra is straightforward with no particularly tricky manipulations. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6x + 2 = 7\sqrt{x} \Rightarrow 6(\sqrt{x})^2 - 7\sqrt{x} + 2 = 0\) | M1 | Expressing as a clear quadratic soi |
| \((3\sqrt{x} - 2)(2\sqrt{x} - 1) = 0\) | M1 | oe e.g. \((3t-2)(2t-1) = 0\) |
| \(\sqrt{x} = \frac{2}{3}\) or \(\frac{1}{2}\) | A1 | 1 solution sufficient. Accept e.g. \(t = 2/3\) |
| \(x = \frac{4}{9}\) or \(\frac{1}{4}\) (or 0.444, 0.25) | A1 | Both solutions required cao |
| OR \((6x+2)^2 = 49x \to 36x^2 - 25x + 4 = 0\) | M1A1 | Attempt to square both sides |
| \((9x-4)(4x-1) = 0\) | M1 | Attempt to solve (or formula etc.) |
| \(x = \frac{4}{9}\) or \(\frac{1}{4}\) (or 0.444, 0.25) oe | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(7^2 - 4 \times 6 \times k (= 0)\) | M1 | Apply \(b^2 - 4ac (= 0)\) |
| \(k = \frac{49}{24}\) or 2.04 | A1 | Attempt to equate derivatives |
| OR \(\frac{d}{dx}(7x^{\frac{1}{3}}) = \frac{d}{dx}(6x + k) \to \frac{7}{2}x^{\frac{-1}{2}} = 6\) | M1 | |
| \(x = \frac{49}{144}\), \(y = \frac{49}{12} \to k = \frac{49}{24}\) or 2.04 | A1 [2] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6x + 2 = 7\sqrt{x} \Rightarrow 6(\sqrt{x})^2 - 7\sqrt{x} + 2 = 0$ | M1 | Expressing as a clear quadratic soi |
| $(3\sqrt{x} - 2)(2\sqrt{x} - 1) = 0$ | M1 | oe e.g. $(3t-2)(2t-1) = 0$ |
| $\sqrt{x} = \frac{2}{3}$ or $\frac{1}{2}$ | A1 | 1 solution sufficient. Accept e.g. $t = 2/3$ |
| $x = \frac{4}{9}$ or $\frac{1}{4}$ (or 0.444, 0.25) | A1 | Both solutions required cao |
| **OR** $(6x+2)^2 = 49x \to 36x^2 - 25x + 4 = 0$ | M1A1 | Attempt to square both sides |
| $(9x-4)(4x-1) = 0$ | M1 | Attempt to solve (or formula etc.) |
| $x = \frac{4}{9}$ or $\frac{1}{4}$ (or 0.444, 0.25) oe | A1 [4] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $7^2 - 4 \times 6 \times k (= 0)$ | M1 | Apply $b^2 - 4ac (= 0)$ |
| $k = \frac{49}{24}$ or 2.04 | A1 | Attempt to equate derivatives |
| **OR** $\frac{d}{dx}(7x^{\frac{1}{3}}) = \frac{d}{dx}(6x + k) \to \frac{7}{2}x^{\frac{-1}{2}} = 6$ | M1 | |
| $x = \frac{49}{144}$, $y = \frac{49}{12} \to k = \frac{49}{24}$ or 2.04 | A1 [2] | |
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\includegraphics[max width=\textwidth, alt={}, center]{4d8fcc3d-a2da-4d98-8500-075d10847be3-2_636_947_1738_598}
The diagram shows the curve $y = 7 \sqrt { } x$ and the line $y = 6 x + k$, where $k$ is a constant. The curve and the line intersect at the points $A$ and $B$.\\
(i) For the case where $k = 2$, find the $x$-coordinates of $A$ and $B$.\\
(ii) Find the value of $k$ for which $y = 6 x + k$ is a tangent to the curve $y = 7 \sqrt { } x$.
\hfill \mbox{\textit{CAIE P1 2012 Q5 [6]}}