oe 3 terms with at least two terms correct condone \(y = x^2 - 10x + 28\)
\(10^2 - 4 \times 1 \times 28\)
M1
oe must be ft from a quadratic derivative
No real roots stated or \(\sqrt{-12}\) or the gradient is never 0
M1dep
dep on both previous M marks must be from correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and not ft
Statement to explain why this means the cubic is always an increasing function or the gradient is always positive
A1
eg gradient can’t be 0 and when \(x = 1\) gradient is 19 so increasing function or graph of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) against \(x\) is always above the \(x\) axis so \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is always positive
\(2[(x - 4)^2 - 4^2] + \ldots\ldots\) or \(2[(x - 4)^2 - 16] + \ldots\ldots\) or \(2\left[(x - 4)^2 - 16 + \dfrac{7}{2}\right]\) or \(2\left[(x - 4)^2 - \dfrac{25}{2}\right]\) or \(2(x - 4)^2 - 2 \times \dfrac{25}{2}\) or \(2(x - 4)^2 - 32 + 7\)
M1dep
oe the bracket is after the \(4^2\) and the 16 here. If they put something else inside the bracket it is incorrect unless it is equivalent to one of the fully complete versions listed
\(2(x - 4)^2 - 25\)
A1
Alternative method 2 – using identities
\(kx^2 + 2kmx + km^2 + n\)
M1
RHS oe
\(k = 2\) and \(2km = -16\) and \(km^2 + n = 7\)
M1dep
oe
\(2(x - 4)^2 - 25\)
A1
Additional guidance
Condone answer followed by \(k = 2 \quad m = -4 \quad n = -25\)
M2A1
\(k = 2 \quad m = -4 \quad n = -25\) without correct answer is penalised
M2A0
Mark scheme (b)
Answer
Mark
Comments
\((x =)\ 1 \pm \sqrt{5}\)
B1
Additional guidance
Condone roots written separately (and/or are both fine)
Do not condone \(1 + \sqrt{5}\) on its own or \(1 - \sqrt{5}\) on its own
19 \(\quad 8x^2 + 20x + n \equiv c(x + d)^2 + 3 \quad\) where \(c\), \(d\) and \(n\) are constants.
Work out the values of \(c\), \(d\) and \(n\). [3 marks]
Mark scheme
Answer
Mark
Comments
(\(c =\)) 8, (\(d =\)) \(\dfrac{5}{4}\) or 1.25 and (\(n =\)) \(\dfrac{31}{2}\) or \(15\dfrac{1}{2}\) or 15.5
B3
oe B2 any 2 correct on answer line (not embedded) B1 (\(d =\)) \(\dfrac{5}{4}\) or 1.25 or (\(n =\)) \(\dfrac{31}{2}\) or \(15\dfrac{1}{2}\) or 15.5 on answer line or \(cx^2 + 2cdx + cd^2\) (+3) or \(8\left(x + \dfrac{5}{4}\right)^2\) or \(8(x + 1.25)^2\) in method
19 Write \(\quad 6x^2 - 24x + 17 \quad\) in the form \(\quad a(x + b)^2 + c \quad\) where \(a\), \(b\) and \(c\) are integers. [3 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\(6(x^2 - 4x \ldots\ldots\) or \(6(x - 2)^2 \ldots\ldots\)
M1
oe eg \(6[(x^2 - 4x) \ldots\ldots]\)
\(6[(x - 2)^2 - 2^2] \ldots\ldots\) or \(6[(x - 2)^2 - 4] \ldots\ldots\) or \(6\left[(x - 2)^2 - 4 + \dfrac{17}{6}\right]\) or \(6\left[(x - 2)^2 - \dfrac{7}{6}\right]\) or \(6(x - 2)^2 - 6 \times \dfrac{7}{6}\) or \(6(x - 2)^2 - 24 + 17\)
M1dep
oe the bracket is after the \(2^2\) and the 4 here. If they put something else inside the bracket it is incorrect unless it is equivalent to one of the fully complete versions listed
13 Prove that \(\quad (3x + 5)^2 - 5x(x + 10) \geqslant 0 \quad\) for all values of \(x\). [4 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\(9x^2 + 15x + 15x + 25 - 5x^2 - 50x\) or \(9x^2 + 30x + 25 - 5x^2 - 50x\) or \(9x^2 + 15x + 15x + 25\) and \(-5x^2 - 50x\) or \(5x^2 + 50x\)
M1
allow only one error in sign, omission or coefficient but not in more than one of these
could be written as 2 separate expansions or in a grid
\(4x^2 - 20x + 25\)
A1
\(4x^2 - 20x + 25\) and \((2x - 5)^2\) or \((2x - 5)(2x - 5)\) or \(4(x - 2.5)^2\) or \(x = 2.5\) or \(b^2 - 4ac = 0\) from quadratic formula
M1dep
factorises or completes the square or uses the quadratic formula correctly. Answer required for M1 dep
\((2x - 5)^2\) or \(4(x - 2.5)^2\) (are squared terms) and so are always \(\geqslant 0\)
A1
oe there must be a stated conclusion eg equal roots and positive quadratic so must be greater than or equal to zero
Alternative method 2
\(9x^2 + 15x + 15x + 25 - 5x^2 - 50x\) or \(9x^2 + 30x + 25 - 5x^2 - 50x\) or \(9x^2 + 15x + 15x + 25\) and \(-5x^2 - 50x\) or \(5x^2 + 50x\)
M1
allow only one error in sign, omission or coefficient but not in more than one of these
could be written as 2 separate expansions or in a grid
\(4x^2 - 20x + 25\)
A1
\(4x^2 - 20x + 25\) and \(\dfrac{\mathrm{d}}{\mathrm{d}x} = 8x - 20\) and is zero when \(x = 2.5\)
M1dep
uses calculus to find stationary point
Tests for minimum by using eg \(x = 2\) and \(x = 3\) or by using 2nd derivative or concludes argument by saying this is a positive quadratic curve with minimum point \((2.5,\ 0)\), hence always \(\geqslant 0\)
\(3(2x - 5)^2 - 70\) or \(a = 3 \quad b = 2 \quad c = -5 \quad d = -70\) or \(3(5 - 2x)^2 - 70\) or \(a = 3 \quad b = -2 \quad c = 5 \quad d = -70\)
A1
oe
Additional guidance
For M marks 2.5 may be seen as \(\dfrac{5}{2}\)
For M marks \((x - 2.5)^2\) may be replaced by \((2.5 - x)^2\) etc
Expansion of given form followed by trial and improvement eg1 \(3(2x - 5)^2 - 70\) (or \(a = 3 \quad b = 2 \quad c = -5 \quad d = -70\)) eg2 Not fully correct
