| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Roots given, find equation constants |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing routine techniques: completing the square (standard drill), finding intersection points by substitution (given one point, finding the other using sum of roots), and finding a line equation between two points. All steps are mechanical with no problem-solving insight required, making it easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02e Complete the square: quadratic polynomials and turning points1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2(x - 1)^2 - 1\) OR \(a = 2, b = -1, c = -1\) | B1, B1, B1 | |
| \(A = (1, -1)\) | B1 ✓ [4] | Allow alt. method for final mark |
| (ii) \(2x^2 - 5x - 3 = 0 \Rightarrow (2x+1)(x-3) = 0\) OE in y | M1, M1 | |
| \(x = -\frac{1}{2}, y = 3\frac{1}{2}\) | A1 [3] | Complete elim & simplify, attempt soln. Additional (3, 7) not penalised |
| (iii) Mid-point of AP = (2, 3) | B1 ✓ | Follow through on their A |
| Gradient of line = \(\frac{1}{2} \div \frac{5}{2} = \frac{-1}{5}\) | B1 | |
| Equation is \(y - 3 = \frac{-1}{5}(x - 2)\) OE | B1 [3] | Or \(y - 3\frac{1}{2} = \frac{-1}{5(x + \frac{1}{2})}\) |
**(i)** $2(x - 1)^2 - 1$ OR $a = 2, b = -1, c = -1$ | B1, B1, B1 |
$A = (1, -1)$ | B1 ✓ [4] | Allow alt. method for final mark
**(ii)** $2x^2 - 5x - 3 = 0 \Rightarrow (2x+1)(x-3) = 0$ OE in y | M1, M1 |
$x = -\frac{1}{2}, y = 3\frac{1}{2}$ | A1 [3] | Complete elim & simplify, attempt soln. Additional (3, 7) not penalised
**(iii)** Mid-point of AP = (2, 3) | B1 ✓ | Follow through on their A
Gradient of line = $\frac{1}{2} \div \frac{5}{2} = \frac{-1}{5}$ | B1 |
Equation is $y - 3 = \frac{-1}{5}(x - 2)$ OE | B1 [3] | Or $y - 3\frac{1}{2} = \frac{-1}{5(x + \frac{1}{2})}$
---10 (i) Express $2 x ^ { 2 } - 4 x + 1$ in the form $a ( x + b ) ^ { 2 } + c$ and hence state the coordinates of the minimum point, $A$, on the curve $y = 2 x ^ { 2 } - 4 x + 1$.
The line $x - y + 4 = 0$ intersects the curve $y = 2 x ^ { 2 } - 4 x + 1$ at points $P$ and $Q$. It is given that the coordinates of $P$ are $( 3,7 )$.\\
(ii) Find the coordinates of $Q$.\\
(iii) Find the equation of the line joining $Q$ to the mid-point of $A P$.
\hfill \mbox{\textit{CAIE P1 2011 Q10 [10]}}