| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Polynomial with line intersection |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring routine algebraic manipulation (substituting line into curve equation, showing discriminant is negative) and basic curve sketching of a quadratic and linear function with axis intercepts. The techniques are standard with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x(5-x) = \tfrac{1}{2}(5x+4)\) | M1 | Forming suitable equation in one variable |
| \(2x^2 - 5x + 4(=0)\) | A1 | Correct 3TQ. Accept \(x^2 - 2.5x + 2(=0)\) etc |
| \(b^2 - 4ac = (-5)^2 - 4\times 2\times 4\) | M1 | Attempt to evaluate discriminant for their 3TQ. Allow \(b^2 > 4ac\) or \(b^2 < 4ac\). Correct formula quoted and some correct substitution |
| \(= 25 - 32 < 0\), so no roots/intersections/solutions | A1 | Correct evaluation for correct 3TQ and comment indicating no roots. Contradictory statements score A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Curve: \(\cap\) shape passing through \((0,0)\) | B1 | Correct shape and passing through origin (can be assumed if passes through intersection of axes) |
| \(\cap\) shape passing through \((5,0)\) | B1 | 5 marked on \(x\)-axis. \(\cap\) shape stopping at both \((5,0)\) and \((0,0)\) scores B0B1 |
| Line: positive gradient and no intersections with \(C\) | B1 | For line of positive gradient that (if extended) has no intersection with their \(C\). Must have both graphs on same axes. If no \(C\) drawn score B0 |
| Line passing through \((0,2)\) and \((-0.8, 0)\) marked on axes | B1 | Accept exact fraction equivalents to \(-0.8\) or \(2\) |
## Question 5:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x(5-x) = \tfrac{1}{2}(5x+4)$ | M1 | Forming suitable equation in one variable |
| $2x^2 - 5x + 4(=0)$ | A1 | Correct 3TQ. Accept $x^2 - 2.5x + 2(=0)$ etc |
| $b^2 - 4ac = (-5)^2 - 4\times 2\times 4$ | M1 | Attempt to evaluate discriminant for their 3TQ. Allow $b^2 > 4ac$ or $b^2 < 4ac$. Correct formula quoted and some correct substitution |
| $= 25 - 32 < 0$, so no roots/intersections/solutions | A1 | Correct evaluation for correct 3TQ and comment indicating no roots. Contradictory statements score A0 |
**ALT:** 2nd M1 for completing the square $a\left[(x\pm\tfrac{b}{2a})^2 - q\right]+c$; 2nd A1 for $(x-\tfrac{5}{4})^2 = -\tfrac{7}{16}$ and suitable comment
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Curve: $\cap$ shape passing through $(0,0)$ | B1 | Correct shape and passing through origin (can be assumed if passes through intersection of axes) |
| $\cap$ shape passing through $(5,0)$ | B1 | 5 marked on $x$-axis. $\cap$ shape stopping at both $(5,0)$ and $(0,0)$ scores B0B1 |
| Line: positive gradient and no intersections with $C$ | B1 | For line of positive gradient that (if extended) has no intersection with their $C$. Must have both graphs on same axes. If no $C$ drawn score B0 |
| Line passing through $(0,2)$ and $(-0.8, 0)$ marked on axes | B1 | Accept exact fraction equivalents to $-0.8$ or $2$ |
---5. The curve $C$ has equation $y = x ( 5 - x )$ and the line $L$ has equation $2 y = 5 x + 4$
\begin{enumerate}[label=(\alph*)]
\item Use algebra to show that $C$ and $L$ do not intersect.
\item In the space on page 11, sketch $C$ and $L$ on the same diagram, showing the coordinates of the points at which $C$ and $L$ meet the axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2012 Q5 [8]}}