| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| 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 standard multi-part C1 question combining quadratic solving, sketching, and finding a tangent condition using the discriminant. Part (iii) requires setting up 3x²-18x-(5+C)=0 and using b²-4ac=0, which is a routine technique taught explicitly in C1. Slightly above average difficulty due to the multi-step nature and tangent condition, but all components are standard textbook exercises. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \((3x+1)(x-5) = 0\) | M1 | Correct method to find roots |
| \(x = -\frac{1}{3}\) or \(x = 5\) | A1, A1 (3) | Correct brackets or formula; both values correct |
| SR B1 | for \(x = 5\) spotted www |
| Answer | Marks |
|---|---|
| B1 | Positive quadratic (must be reasonably symmetrical) |
| B1 | \(y\)-intercept correct |
| B1ft (3) | both \(x\)-intercepts correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 6x - 14\), \(6x - 14 = 4\), \(x = 3\) | M1*, M1*, A1 | Use differentiation to find gradient; equating gradient expression to 4; \(x\) value |
| On curve when \(x = 3\), \(y = -20\) | A1ft | Finding \(y\) co-ordinate for their \(x\) value |
| \(-20 = (4 \times 3) + c\), \(c = -32\) | M1dep, A1 (6) | Dependent on both previous M marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x^2 - 14x - 5 = 4x + c\) | M1 | Equate curve and line |
| \(3x^2 - 18x - 5 - c = 0\) has one solution | B1 | Statement that only one solution for tangent |
| \(b^2 - 4ac = 0\) | M1 | Use of discriminant \(= 0\) |
| \((-18)^2 - (4 \times 3 \times (-5-c)) = 0\) | M1 | Attempt to use \(a, b, c\) from their equation |
| \(c = -32\) | A1, A1 (6) |
# Question 10:
**(i)**
$(3x+1)(x-5) = 0$ | M1 | Correct method to find roots
$x = -\frac{1}{3}$ or $x = 5$ | A1, A1 (3) | Correct brackets or formula; both values correct
| SR B1 | for $x = 5$ spotted **www**
**(ii)**
| B1 | Positive quadratic (must be reasonably symmetrical)
| B1 | $y$-intercept correct
| B1ft (3) | both $x$-intercepts correct
**(iii)**
$\frac{dy}{dx} = 6x - 14$, $6x - 14 = 4$, $x = 3$ | M1*, M1*, A1 | Use differentiation to find gradient; equating gradient expression to 4; $x$ value
On curve when $x = 3$, $y = -20$ | A1ft | Finding $y$ co-ordinate for their $x$ value
$-20 = (4 \times 3) + c$, $c = -32$ | M1dep, A1 (6) | Dependent on both previous M marks
**Alternative method:**
$3x^2 - 14x - 5 = 4x + c$ | M1 | Equate curve and line
$3x^2 - 18x - 5 - c = 0$ has one solution | B1 | Statement that only one solution for tangent
$b^2 - 4ac = 0$ | M1 | Use of discriminant $= 0$
$(-18)^2 - (4 \times 3 \times (-5-c)) = 0$ | M1 | Attempt to use $a, b, c$ from their equation
$c = -32$ | A1, A1 (6) |10 (i) Solve the equation $3 x ^ { 2 } - 14 x - 5 = 0$.
A curve has equation $\mathrm { y } = 3 \mathrm { x } ^ { 2 } - 14 \mathrm { x } - 5$.\\
(ii) Sketch the curve, indicating the coordinates of all intercepts with the axes.\\
(iii) Find the value of C for which the line $\mathrm { y } = 4 \mathrm { x } + \mathrm { C }$ is a tangent to the curve.
\hfill \mbox{\textit{OCR C1 2007 Q10 [12]}}