| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show line is tangent, verify |
| Difficulty | Moderate -0.8 This is a routine C1 question testing standard completing-the-square technique, basic curve sketching, and simultaneous equations. Part (iv) requires recognizing that one solution means tangency, but this is a well-practiced concept. All parts follow textbook procedures with no novel problem-solving 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 points1.02n Sketch curves: simple equations including polynomials1.02p Interpret algebraic solutions: graphically |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 - 7\) | 3 | mark final answer; 1 for \(a=3\), 2 for \(b=7\) or M1 for \(-3^2+2\); bod 3 for \((x-3)-7\) |
| Answer | Marks |
|---|---|
| \((3, -7)\) or ft from (i) | \(1+1\) |
| Answer | Marks | Guidance |
|---|---|---|
| sketch of quadratic correct way up and through \((0, 2)\) | G1 | accept \((0,2)\) o.e. seen in this part |
| t.p. correct or ft from (ii) | G1 | accept 3 and \(-7\) marked on axes level with turning pt., or better; no ft for \((0,2)\) as min |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 6x + 2 = 2x - 14\) o.e. | M1 | or their (i) \(= 2x - 14\) |
| \(x^2 - 8x + 16 [=0]\) | M1 | dep on first M1; condone one error |
| \((x-4)^2 [=0]\) | M1 | or correct use of formula giving equal roots; allow \((x+4)^2\) o.e. ft \(x^2 + 8x + 16\) |
| \(x = 4\), \(y = -6\) | A1 | if M0M0M0, allow SC2 for showing \((4,-6)\) is on both graphs |
| equal/repeated roots implies tangent — must be explicitly stated | A1 | or for use of calculus to show grad of line and curve are same when \(x=4\) |
# Question 10:
## Part (i)
| $(x-3)^2 - 7$ | 3 | mark final answer; 1 for $a=3$, 2 for $b=7$ or M1 for $-3^2+2$; bod 3 for $(x-3)-7$ |
## Part (ii)
| $(3, -7)$ or ft from (i) | $1+1$ | |
## Part (iii)
| sketch of quadratic correct way up and through $(0, 2)$ | G1 | accept $(0,2)$ o.e. seen in this part |
| t.p. correct or ft from (ii) | G1 | accept 3 and $-7$ marked on axes level with turning pt., or better; no ft for $(0,2)$ as min |
## Part (iv)
| $x^2 - 6x + 2 = 2x - 14$ o.e. | M1 | or their (i) $= 2x - 14$ |
| $x^2 - 8x + 16 [=0]$ | M1 | dep on first M1; condone one error |
| $(x-4)^2 [=0]$ | M1 | or correct use of formula giving equal roots; allow $(x+4)^2$ o.e. ft $x^2 + 8x + 16$ |
| $x = 4$, $y = -6$ | A1 | if M0M0M0, allow SC2 for showing $(4,-6)$ is on both graphs |
| equal/repeated roots implies tangent — must be explicitly stated | A1 | or for use of calculus to show grad of line and curve are same when $x=4$ |
---10 (i) Express $x ^ { 2 } - 6 x + 2$ in the form $( x - a ) ^ { 2 } - b$.\\
(ii) State the coordinates of the turning point on the graph of $y = x ^ { 2 } - 6 x + 2$.\\
(iii) Sketch the graph of $y = x ^ { 2 } - 6 x + 2$. You need not state the coordinates of the points where the graph intersects the $x$-axis.\\
(iv) Solve the simultaneous equations $y = x ^ { 2 } - 6 x + 2$ and $y = 2 x - 14$. Hence show that the line $y = 2 x - 14$ is a tangent to the curve $y = x ^ { 2 } - 6 x + 2$.
\hfill \mbox{\textit{OCR MEI C1 2008 Q10 [12]}}