Edexcel C1 — Question 4 7 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks7
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TopicDiscriminant and conditions for roots
TypeLine tangent to curve, find k for tangency
DifficultyStandard +0.3 Part (a) is routine quadratic formula application. Part (b) requires setting up a discriminant condition (b²-4ac=0) for tangency, which is a standard C1 technique but slightly beyond pure recall. The multi-step nature and tangency concept elevate this slightly above average difficulty.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations
4. (a) Find in exact form the coordinates of the points where the curve \(y = x ^ { 2 } - 4 x + 2\) crosses the \(x\)-axis.
(b) Find the value of the constant \(k\) for which the straight line \(y = 2 x + k\) is a tangent to the curve \(y = x ^ { 2 } - 4 x + 2\).
AnswerMarks Guidance
(a) \(x^2 - 4x + 2 = 0\)M2
\(x = \frac{4 \pm \sqrt{16-8}}{2} = \frac{4 \pm 2\sqrt{2}}{2}\)
\(x = 2 \pm \sqrt{2}\), so \((2 - \sqrt{2}, 0), (2 + \sqrt{2}, 0)\)A2
(b) For tangency: \(x^2 - 4x + 2 = 2x + k\) gives \(x^2 - 6x + 2 - k = 0\)M1 A1
Equal roots require \(b^2 - 4ac = 0\): \((-6)^2 - 4(1)(2-k) = 0\)M1 A1
\(36 - 4(2 - k) = 0\), so \(k = -7\)A1 (7) 
**(a)** $x^2 - 4x + 2 = 0$ | M2 |
$x = \frac{4 \pm \sqrt{16-8}}{2} = \frac{4 \pm 2\sqrt{2}}{2}$ |  |
$x = 2 \pm \sqrt{2}$, so $(2 - \sqrt{2}, 0), (2 + \sqrt{2}, 0)$ | A2 |

**(b)** For tangency: $x^2 - 4x + 2 = 2x + k$ gives $x^2 - 6x + 2 - k = 0$ | M1 A1 |
Equal roots require $b^2 - 4ac = 0$: $(-6)^2 - 4(1)(2-k) = 0$ | M1 A1 |
$36 - 4(2 - k) = 0$, so $k = -7$ | A1 | (7)
4. (a) Find in exact form the coordinates of the points where the curve $y = x ^ { 2 } - 4 x + 2$ crosses the $x$-axis.\\
(b) Find the value of the constant $k$ for which the straight line $y = 2 x + k$ is a tangent to the curve $y = x ^ { 2 } - 4 x + 2$.\\

\hfill \mbox{\textit{Edexcel C1  Q4 [7]}}
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