Higher June 2018 Paper 2 Q26
26 A curve has equation \(\quad y = 4x^2 + 5x + 3\)
A line has equation \(\quad y = x + 2\)
Show that the curve and the line have exactly one point of intersection.
Do not use a graphical method. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(4x^2 + 5x + 3 = x + 2\) | M1 | |
| \(4x^2 + 5x - x + 3 - 2\) (\(= 0\)) or \(4x^2 + 4x + 1\) (\(= 0\)) | M1dep | oe collection of terms eg \(4x^2 + 5x - x = 2 - 3\) or \(4x^2 + 4x = -1\) |
| \((2x + 1)(2x + 1)\) (\(= 0\)) or \(4\left(x + \dfrac{1}{2}\right)^2\) (\(= 0\)) or \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times 1}}{2 \times 4}\) or \(b^2 - 4ac = 4^2 - 4 \times 4 \times 1\) or D(iscriminant) \(= 4^2 - 4 \times 4 \times 1\) | A1 | oe eg \(\left(x + \dfrac{1}{2}\right)^2\) (\(= 0\)) allow \(b^2 - 4ac = 16 - 16\) or D(iscriminant) \(= 16 - 16\) |
| (\(x =\)) \(-\dfrac{1}{2}\) with no other solutions with M2A1 seen or states that as brackets are the same there is only one solution with M2A1 seen or \(b^2 - 4ac = 4^2 - 4 \times 4 \times 1 = 0\) and states there is only one solution with M2A1 seen or D(iscriminant) \(= 4^2 - 4 \times 4 \times 1 = 0\) and states there is only one solution with M2A1 seen | A1 | oe allow \(b^2 - 4ac = 16 - 16 = 0\) and states there is only one solution with M2A1 seen allow D(iscriminant) \(= 16 - 16 = 0\) and states there is only one solution with M2A1 seen |
| Alternative method 2 | ||
| \(y = 4(y - 2)^2 + 5(y - 2) + 3\) | M1 | oe |
| \(4y^2 - 16y + 16 + 5y - 10 + 3 - y\) (\(= 0\)) or \(4y^2 - 12y + 9\) (\(= 0\)) | M1dep | oe expansion and collection of terms eg \(4y^2 - 16y + 5y - y = 10 - 16 - 3\) or \(4y^2 - 12y = -9\) |
| \((2y - 3)(2y - 3)\) (\(= 0\)) or \(4\left(y - \dfrac{3}{2}\right)^2\) (\(= 0\)) or \(\dfrac{--12 \pm \sqrt{(-12)^2 - 4 \times 4 \times 9}}{2 \times 4}\) or \(b^2 - 4ac = (-12)^2 - 4 \times 4 \times 9\) or D(iscriminant) \(= (-12)^2 - 4 \times 4 \times 9\) | A1 | oe eg \(\left(y - \dfrac{3}{2}\right)^2\) (\(= 0\)) allow \(b^2 - 4ac = 144 - 144\) or allow D(iscriminant) \(= 144 - 144\) |
| (\(y =\)) \(\dfrac{3}{2}\) with no other solutions with M2A1 seen or states that as brackets are the same there is only one solution with M2A1 seen or \(b^2 - 4ac = (-12)^2 - 4 \times 4 \times 9 = 0\) and states there is only one solution with M2A1 seen or D(iscriminant) \(= (-12)^2 - 4 \times 4 \times 9 = 0\) and states there is only one solution with M2A1 seen | A1 | oe allow \(b^2 - 4ac = 144 - 144 = 0\) and states there is only one solution with M2A1 seen allow D(iscriminant) \(= 144 - 144 = 0\) and states there is only one solution with M2A1 seen |
Additional guidance
| Alt 1 (\(x =\)) \(-\dfrac{1}{2}\) with no working or Alt 2 (\(y =\)) \(\dfrac{3}{2}\) with no working | M0M0A0A0 |
| Alt 1 Ignore any \(y\)-coordinate whether correct \(\left(= \dfrac{3}{2}\right)\) or incorrect | |
| Alt 2 Ignore any \(x\)-coordinate whether correct \(\left(= -\dfrac{1}{2}\right)\) or incorrect | |
| T & I leading to \(x = -\dfrac{1}{2}\) | M0M0A0A0 |
| To award M1dep you must see a correct expression with terms collected or a correct equation with terms collected | |
| \(4x^2 + 5x + 3 = x + 2\) \(4x^2 + 1 = -4x\) (all \(x\) terms not collected on one side) | M1 M0dep |
| \(4x^2 + 5x + 3 = x + 2\) \(4x^2 + 4x + 3 = 2\) (all constant terms not collected on one side) | M1 M0dep |
| If using the discriminant to award A marks, you must see either \(b^2 - 4ac\) or D(iscriminant) \(b^2 - 4ac = 4^2 - 4 \times 4 \times 1\) can be implied eg \(b + \sqrt{b^2 - 4ac}\) and \(4 + \sqrt{4^2 - 4 \times 4 \times 1}\) scores first A1 For final A1 must see \(b^2 - 4ac = 4^2 - 4 \times 4 \times 1 = 0\) and statement that there is only one solution with M2A1 seen |