20 Here is the graph of \(\quad y = -x^2 + 3x + 6 \quad\) for \(\quad -2 \leqslant x \leqslant 5\)
By drawing a suitable straight line, work out approximate solutions to
\[x^2 - \frac{7}{2}x - 3 = 0\]
[4 marks]
Mark scheme
Answer
Mark
Comments
\((y =)\ \dfrac{-x}{2} + 3\)
B2
oe eg \(2y = -x + 6\) can be implied from correct line B1 for \((y =)\ \dfrac{x}{2} - 3\) can be implied from drawing this line or \(m = -\dfrac{1}{2}\) or \(c = +3\) not from incorrect working for either \(m\) or \(c\)
\(y = \dfrac{-x}{2} + 3\) accurately drawn on diagram
M1
ft their stated linear function in \(x\) accuracy is half a square at \(x = 0\) and \(x = 4\) needs to be a straight line (mark intention)
\([-0.6, -0.8]\) and [4.1, 4.3]
A1
dependent on B2M1
Additional guidance
Factorising by use of the quadratic formula alone
B0M0A0
Ignore any y values for A1
Do not allow M mark for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) linear function drawn
7 Here is a sketch of the curve \(\quad y = ax^2 + bx + c \quad\) where \(a\), \(b\) and \(c\) are constants.
The curve intersects the \(x\)-axis at \((-4, 0)\) and \((p, 0)\)
The turning point has \(x\)-coordinate 0.5
(a) Work out the value of \(p\). [1 mark]
(b) Solve \(\quad ax^2 + bx + c \gt 0\) [2 marks]
Mark scheme (a)
Answer
Mark
Comments
5
B1
condone (5, 0) but not (0, 5)
Additional guidance
Condone 5 or (5, 0) next to \(p\) on sketch if answer line blank
B1
Mark scheme (b)
Answer
Mark
Comments
Either of \(x \lt -4 \qquad x \gt \text{their } 5\)
M1
oe eg \(-4 \gt x\) correct or ft their 5 in (a) their 5 in (a) must be greater than 0.5 condone \(\leqslant\) for \(\lt\) condone \(\geqslant\) for \(\gt\)
Both of \(x \lt -4 \qquad x \gt \text{their } 5\)
A1ft
correct or ft their 5 in (a) their 5 in (a) must be greater than 0.5 must be two separate inequalities
Additional guidance
Ignore the use of or/and eg1 \(x \lt -4\) or \(x \gt 5\) eg2 \(x \lt -4\) and \(x \gt 5\)
M1A1 M1A1
Condone embedded even if in an invalid statement for M1 but not A1 eg1 \(-4 \gt x \gt 5\) eg2 \(-4 \gt x \gt 6\)
M1A0 M1A0
If answer to (a) is 4 \(x \lt -4 \qquad x \gt 4\)
M1A1ft
If answer to (a) is blank eg1 \(x \lt -4\) eg2 \(x \lt -4 \qquad x \gt 5\)
M1A0 M1A1
\(-4 \lt x \lt 5\)
M0A0
If answer to (a) is eg (0, 4) take their 5 to be 4
Condone eg \(x = \lt -4\) for M1 but not A1
Allow eg \((-\infty, -4)\) for \(x \lt -4\) (condone eg \([-\infty, -4)\) for M1 but not A1)
(a) \[\begin{aligned} \mathrm{f}(x) &= 4 - x &&\quad 0 \leqslant x \lt 1 \\ &= 4x - x^2 &&\quad 1 \leqslant x \lt 4 \\ &= 2x - 8 &&\quad 4 \leqslant x \leqslant 6 \end{aligned}\]
On the grid, draw the graph of \(\quad y = \mathrm{f}(x)\) [4 marks]
(b) \(\mathrm{g}(x) = 6 - 3x\)
Work out \(\;\mathrm{g}^{-1}(x)\). [2 marks]
Mark scheme (a)
Answer
Mark
Comments
Line joining (0,4) and (1,3)
B1
may be drawn free hand
(2, 4) plotted as a maximum value
M1
needs to be some sort of graph showing a maximum value
Curve drawn through (1, 3), (2, 4), (3, 3) and (4, 0)
A1
all points should be within half a square horizontally or vertically
Line joining (4,0) and (6,4)
B1
may be drawn free hand
Additional guidance
Maximum mark available if either or both of the straight lines include a curve (they haven’t used a ruler) is B1M1A1 Maximum mark available if any part of the quadratic curve (it must be a quadratic curve that is concave and not convex at any point) is drawn with a ruler is B2M1 (a clear vertex at (3,3) may show the use of a ruler) Maximum mark available if any of the lines go beyond their correct domain by more than half a square is B2M1 or B1M1A1 They could lose marks for both the quadratic and straight lines Ignore slight feathering
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1
Rearranging first to get \(x = \dfrac{6 - \mathrm{g}(x)}{3}\)
M1
oe eg \(x = \dfrac{6 - y}{3}\) or \(2 - \dfrac{y}{3}\) \(y - 6 = -3x\) is not enough to gain M1
\(\mathrm{g}^{-1}(x) = \dfrac{6 - x}{3}\)
A1
oe eg \(\mathrm{g}^{-1}(x) = \dfrac{x - 6}{-3}\) or \(\mathrm{g}^{-1}(x) = -\dfrac{x - 6}{3}\) or \(\mathrm{g}^{-1}(x) = \dfrac{x}{-3} + 2\)
Alternative method 2
Putting the correct terminology in to get \(\;x = 6 - 3\mathrm{g}^{-1}(x)\)
M1
oe eg \(x = 6 - 3y\) or \(3y = 6 - x\)
\(\mathrm{g}^{-1}(x) = \dfrac{6 - x}{3}\)
A1
oe eg \(\mathrm{g}^{-1}(x) = \dfrac{x - 6}{-3}\) or \(\mathrm{g}^{-1}(x) = -\dfrac{x - 6}{3}\) or \(\mathrm{g}^{-1}(x) = \dfrac{x}{-3} + 2\)
Additional guidance
Answer left as \(y = \dfrac{6 - x}{3}\) should gain M1 on either scheme
M1A0
\(x = \dfrac{6 - y}{3}\) can gain M1 but not A1
M1A0
Condone \(\mathrm{g}^{-1}(x)\) missed on answer line (as long as nothing else is written in its place) Flow charts may be used. Mark as oe Penalise additional incorrect working
19 \(y = \mathrm{f}(x) \quad\) is the graph of a cubic function.
\(y \lt 0 \quad\) for \(\quad x \lt 5\) \(y \geqslant 0 \quad\) for \(\quad x \geqslant 5\)
The function is
increasing for \(\quad x \lt -1\) decreasing for \(\quad -1 \lt x \lt 2\) increasing for \(\quad x \gt 2\)
Draw a possible sketch of \(\quad y = \mathrm{f}(x) \quad\) for values of \(x\) from \(-2\) to 6 [4 marks]
Mark scheme
Answer
Mark
Comments
Cubic curve from \(x = -2\) to \(x = 6\) and maximum point at \((-1, a)\) where \(a\) is negative and minimum point at \((2, b)\) where \(b\) is less than \(a\) and increasing through (5, 0)
B4
B3 curve from \(x = -2\) to \(x = 6\) and maximum point at \((-1, c)\) where \(c\) is any value and minimum point at \((2, d)\) where \(d\) is less than \(c\) and \(d\) is negative and increasing through (5, 0) or a B4 response apart from cubic curve not drawn from \(x = -2\) to \(x = 6\)
B2 curve with maximum point at \((-1, e)\) where \(e\) is any value and minimum point at \((2, f)\) where \(f\) is less than \(e\)
B1 curve with maximum point at \((-1, g)\) where \(g\) is negative or curve with minimum point at \((2, h)\) where \(h\) is negative or curve increasing through (5, 0) SC2 max and min correct and increasing through (5, 0) but with straight lines rather than a curve.
\[\begin{aligned} \mathrm{f}(x) &= 4x && x \lt 0 \\ &= x^2 - 8x && 0 \leqslant x \leqslant 8 \\ &= 16 - 2x && x \gt 8 \end{aligned}\]
A sketch of \(\quad y = \mathrm{f}(x) \quad\) is shown.
Not drawn accurately
Work out all the values of \(x\) for which \(\quad \mathrm{f}(x) = -12\) [4 marks]
Mark scheme
Answer
Mark
Comments
\(-3 \quad 2 \quad 6 \quad 14\) with no other solutions
B4
B3 three correct with at most one incorrect B2 two correct with at most two incorrect B1 one correct with at most three incorrect SC2 \(-3 \quad 2 \quad 6 \quad 14\) with no other values seen SC1 Two or three of \(-3 \quad 2 \quad 6 \quad 14\) with no other values seen
Additional guidance
Solutions may be in any order eg1 \(-3 \quad 14 \quad 6 \quad 2\) eg2 \(14 \quad -3\)
B4 B2
\(x \lt -3 \quad 2 \lt x \lt 6 \quad x \gt 14\)
SC2
\(2 \leqslant x \leqslant 6\)
SC1
\(-3 \quad 2 \quad 6 \quad 14\) seen in working with no other values and answer line \(-3 \leqslant x \leqslant 14\)
has a maximum point at \(L\ (-1, 13)\) has a minimum point at \(M\ (2, -14)\) intersects the \(y\)-axis at \(N\).
The curve crosses the \(x\)-axis at three distinct points.
On the axes below, sketch the curve.
Label the points \(L\), \(M\) and \(N\) on your sketch. [3 marks]
Mark scheme
Answer
Mark
Comments
Correct shape curve crossing \(x\)-axis twice for \(x \gt 0\) crossing \(x\)-axis once for \(x \lt 0\) maximum point \(L\) labelled minimum point \(M\) labelled \(N\ (0, 6)\) labelled
B3
B2 Correct shape curve crossing \(x\)-axis twice for \(x \gt 0\) crossing \(x\)-axis once for \(x \lt 0\) with incomplete labelling B1 Identifies \(N\) as (0, 6)
Draw the graph of \(\quad y = \mathrm{f}(x) \quad\) on the grid. [4 marks]
Mark scheme
Answer
Mark
Comments
B4
B1 for a straight line from \((-1,\ 2)\) to \((0,\ 0)\). This should be drawn with a ruler (give BOD)
B1 for a quadratic style curve through \((0,\ 0)\), \((1,\ 3)\), \((2,\ 4)\) and \((3,\ 3)\) within tolerance. Condone one straight line between only one of the sections between \((0,\ 0)\) and \((1,\ 3)\), \((1,\ 3)\) and \((2,\ 4)\) or \((2,\ 4)\) and \((3,\ 3)\)
B1 for any quadratic graph drawn with correct curvature and no straight lines (see examples) with clear vertices at \((0,0)\) and \((3,3)\) and within tolerance for these points. There needs to be evidence of a maximum turning point drawn.
B1 for a straight line from \((3,\ 3)\) to \((4,\ 5)\). This should be drawn with a ruler (give BOD)
SC1 for all six stated points plotted correctly and clearly defined (don’t need to be joined up or could be joined up incorrectly). If any incorrect points plotted then no marks can be awarded
Additional guidance
Tolerance of plot \(\pm\) 2mm for each stated point (these are 1cm squares)
For \(\mathrm{f}(x) = -2x\) extending to the left of \(x = -1\) or \(\mathrm{f}(x) = 2x - 3\) extending to the right of \(x = 4\) greater than 2mm, award maximum 1 mark from B1 B1 only for a section that would have otherwise scored ie. If both lines extended then B0 B1
If more than one line drawn in any section then choice so loss of marks
Ignore shading under or over the lines as long as the graph is clear
This would score for the second B mark only
Some feathering but close enough to score B4
B0 but had the line stopped at (4,5) it would have gained B1
Straight lines both correct but penalised one mark for the first one extending beyond the domain B1. Quadratic out of tolerance for (1,3) B0. Correct curvature and vertices B1
The \(y\)-coordinate of the minimum point is \(2\dfrac{1}{3}\) The \(y\)-coordinate of the maximum point is 13
\((0,\ 4)\) is a point on the curve. The tangent at \((0,\ 4)\) has a negative gradient.
Sketch the curve on the grid below. Show the coordinates of the stationary points. [4 marks]
Mark scheme
Answer
Mark
Comments
factorising to get \((x + 3)(x - 1)\ (= 0)\) or completing the square and getting as far as \(x + 1 = \pm 2\) or using the quadratic formula and getting as far as \(x = \dfrac{-2 \pm 4}{2}\)
M1
\(x = -3\) and \(x = 1\)
A1
\((-3,\ 13)\) as a maximum point and \((1,\ 2\tfrac{1}{3})\) as a minimum point, plotted
M1
Smooth correct curve which must have the stationary points plotted in the correct quadrants and must cross the negative \(x\)-axis
A1
Additional guidance
SC1 for a fully correct sketch with the stationary points in the correct quadrants but lacking any detail in terms of the \(x\) coordinates of the stationary points, or with incorrect values of the stationary points, and with no evidence of a valid method to obtain \(x = -3\) and \(x = 1\)
15 Here is the graph of \(\quad y = 3x - x^2 \quad\) for values of \(x\) from \(-1\) to 4
By drawing a suitable linear graph on the grid, work out approximate solutions to
\[x^2 - 4x + 2 = 0\]
[4 marks]
Mark scheme
Answer
Mark
Comments
\(2 - x\) or \(x - 2\)
M1
do not award M1 if you see evidence of incorrect method for finding a linear expression
\(y = 2 - x\) accurately drawn
M1
3.4
A1
accept 3.3 to 3.5
0.6
A1
accept 0.5 to 0.7
Additional guidance
For the first M1, start by looking for evidence of a correct method. eg \(\ x^2 - 4x + 2 + 3x - x^2 = -x + 2\) or \(x^2 - 4x + 2 = 0 \ \rightarrow\ x^2 - 3x - x + 2 = 0 \ \rightarrow\ -x + 2 = 3x - x^2\)
Attempts to solve \(x^2 - 4x + 2 = 0\) by using the quadratic formula or by completing the square or by drawing a new quadratic graph (for \(y = x^2 - 4x + 2\)) score 0 marks You might see work which uses the quadratic formula or completing the square which leads to answers of \(2 \pm \sqrt{2}\) … and if this follows working using a correct method to find the linear graph, it can be ignored (they could be using it as a check on their answers obtained graphically), but if it looks like it is their main method, then award 0 marks, as stated above.. Ignore any \(y\) coordinates that might accompany the final \(x\) values.
5 On the grid, draw the graph of \(\quad y = \mathrm{f}(x)\)
\[\begin{aligned} \mathrm{f}(x) &= x + 4 &\qquad& -4 \leqslant x \lt 0 \\ &= 4 - 3x && 0 \leqslant x \lt 2 \\ &= -2 && 2 \leqslant x \leqslant 5 \end{aligned}\]
[4 marks]
Mark scheme
Answer
Mark
Comments
Line from \((-4,\ 0)\) to \((0,\ 4)\)
M1
mark intention
Line from \((0,\ 4)\) to \((2,\ -2)\)
M1
lines do not have to be straight but must pass through all integer points
Line from \((2,\ -2)\) to \((5,\ -2)\)
M1
only condone the first instance of a line that extends beyond the given domain
Straight line from \((-4,\ 0)\) to \((0,\ 4)\) and straight line from \((0,\ 4)\) to \((2,\ -2)\) and straight line from \((2,\ -2)\) to \((5,\ -2)\)
A1
all straight lines must be the correct length with no other lines graph must be accurate SC3 \((-4,\ 0)\) and \((-3,\ 1)\) and \((-2,\ 2)\) and \((-1,\ 3)\) and \((0,\ 4)\) and \((1,\ 1)\) and \((2,\ -2)\) and \((3,\ -2)\) and \((4,\ -2)\) and \((5,\ -2)\) plotted (any other points plotted must be correct ones for the graph) SC2 \((-4,\ 0)\) and \((0,\ 4)\) and \((2,\ -2)\) and \((5,\ -2)\) plotted (any other points plotted must be correct ones for the graph)
Additional guidance
(crosses do not have to be shown)
M3A1
Dashed or dotted lines can score up to M3A0
Points may be implied by a correct line
M mark examples eg1 2 correct lines and 1 extended line (but otherwise correct) eg2 1 correct line and 2 extended lines (but otherwise correct) eg3 3 extended lines (but otherwise correct)
21 Here is a sketch of \(\quad y = (x + 2)(4 - x)\)
The graph intersects the axes at \(A\) (\(-2\), 0), \(B\) (4, 0) and \(C\).
Not drawn accurately
(a) Work out the coordinates of \(C\). [1 mark]
(b) Work out the gradient function of the curve. [3 marks]
(c) The normal to the curve at \(C\) intersects the \(x\)-axis at \(D\).
Show that \(\qquad\) length \(BD = 2 \times\) length \(AB\) [5 marks]
Mark scheme (a)
Answer
Mark
Comments
(0, 8)
B1
Additional guidance
Answer line takes precedence over working lines and diagram
Answer line blank with \(C\) labelled (0, 8) on diagram
B1
Answer line blank with 8 written next to \(C\) on diagram
B0
(8, 0)
B0
Answer 8
B0
Mark scheme (b)
Answer
Mark
Comments
\(-x^2 - 2x + 4x + 8\)
M1
Allow one error but no omission Must have an \(x^2\) term Terms may be seen in a grid Implied by \(-x^2 + 2x + k \quad k \neq 0\) or \(ax^2 + 2x + 8 \quad a \neq 0\)
\(-x^2 - 2x + 4x + 8\) or \(-x^2 + 2x + 8\)
A1
\(-x^2 - 2x + 4x + 8\) but an error in any collection of terms is M1A0
\(-2x - 2 + 4\) or \(-2x + 2\) or \(-2(x - 1)\) or \(2(1 - x)\)
A1ft
oe ft their quadratic in \(x\) with M1 awarded
Additional guidance
\(2 - 2x\) with final answer 2 (from substituting in \(x = 0\))
M1A1A0
Condone \(y = 2 - 2x\) or \(\mathrm{f}(x) = 2 - 2x\) in working for M1A1 If (\(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\mathrm{f}'(x)\) =) \(2 - 2x\) on answer line also award final A1
\(y = 2 - 2x\) or \(\mathrm{f}(x) = 2 - 2x\) on answer line
M1A1A0
When marking (b), a maximum of M1A1A0 can be awarded from an expansion seen on the previous page if not contradicted by an expansion in (b) The final A1 must be seen in (b) eg1 (b) no expansion seen with an answer of \(2x + 2\) At top of previous page \(\ -x^2 + 2x + 8\) eg2 (b) no expansion seen with an answer of \(-2x + 6\) In (a) \(\ -x^2 + 2x + 4x + 8 = -x^2 + 6x + 8\)
M1A1A0
M1A0A1ft
Correct use of product rule and gradient function \(= -2x + 2\)
3 marks
Mark scheme (c)
Answer
Mark
Comments
Alternative method 1
(gradient of curve at \(C\) =) 2
B1ft
Correct or ft their (b) when \(x = 0\) May be implied
\(-\dfrac{1}{\text{their } 2}\) or \(-\dfrac{1}{2}\)
M1
oe ft their 2 Only ft a non-zero numerical value (gradient of normal =) \(-\dfrac{1}{2}\) is B1M1
\(y = \left(\text{their } -\dfrac{1}{2}\right)x +\) their 8 or \(y -\) their \(8 =\) their \(-\dfrac{1}{2}(x - 0)\)
M1dep
Must have used gradient of normal not gradient of tangent Correct or ft their 8 from (a) in the form (0, \(k\))
\(0 = \left(\text{their } -\dfrac{1}{2}\right)x +\) their 8 or \(0 -\) their \(8 =\) their \(-\dfrac{1}{2}(x - 0)\) or \(x = 16\)
M1dep
Correct or ft their 8 from (a) in the form (0, \(k\))
\(x = 16\) and \(BD = 12\) and \(AB = 6\) with correct method seen
A1
oe
Alternative method 2
(gradient of curve at \(C\) =) 2
B1ft
Correct or ft their (b) when \(x = 0\) May be implied
\(-\dfrac{1}{\text{their } 2}\) or \(-\dfrac{1}{2}\)
M1
oe ft their 2 Only ft a non-zero numerical value (gradient of normal =) \(-\dfrac{1}{2}\) is B1M1
\(\dfrac{0 - \text{their } 8}{x - 0} =\) their \(-\dfrac{1}{2}\) or \(x = -\)their \(8 \div\) their \(-\dfrac{1}{2}\) or \(x = 16\)
M2dep
oe Correct or ft their 8 from (a) in the form (0, \(k\))
\(x = 16\) and \(BD = 12\) and \(AB = 6\) with correct method seen
(a) Draw the graph of \(\quad y = \mathrm{f}(x) \quad\) on the axes below. [4 marks]
(b) State the range of \(\ \mathrm{f}(x)\) [2 marks]
Mark scheme (a)
Answer
Mark
Comments
Straight line between (\(-2\), 7) and (0, 3)
B1
Tolerance of ±1 small square Allow line to be extended
Points (0, 3) (1, 4) (2, 3) (3, 0) (4, \(-5\))
M1
Tolerance of ±1 small square May be plotted or seen in a table Points can be implied
Correct smooth parabolic curve with maximum at (1, 4)
A1
Tolerance of ±1 small square Allow (ruled) straight line between (3, 0) and (4, \(-5\)) Curve passing through all correct points within tolerance scores M1A1
Straight line between (4, \(-5\)) and (5, 0)
B1
Tolerance of ±1 small square Allow line to be extended
Additional guidance
Ignore extra points plotted
Tolerance of ±1 small square means it is on the edges of or within the shaded area
Points only can score a maximum of M1
Ruled straight lines for curve apart from between (3, 0) and (4, \(-5\))
A0
If all 4 marks would be awarded but either (i) graph has a line or a curve that extends beyond the individual domains or (ii) the curve does not meet a line at a cusp
3 marks
Mark scheme (b)
Answer
Mark
Comments
\(-5 \leqslant \mathrm{f}(x) \leqslant 7\) or \(7 \geqslant \mathrm{f}(x) \geqslant -5\) or [\(-5\), 7]
B2ft
Correct or ft their graph in (a) for B2 ft their graph in (a) for B1
B1ft \(-5 \leqslant \mathrm{f}(x)\) or \(\mathrm{f}(x) \leqslant 7\) on their own or embedded within an interval for \(\mathrm{f}(x)\) or only \(-5\) and 7 chosen eg \(-5 \lt \mathrm{f}(x) \lt 7\)
Additional guidance
Allow \(\mathrm{f}(x)\) to be \(y\) or f or f\(x\) eg1 \(-5 \leqslant y \leqslant 7\) eg2 \(\mathrm{f} \leqslant 7\)
B2 B1
Allow as two inequalities \(\quad \mathrm{f}(x) \geqslant -5\) (and/or) \(\mathrm{f}(x) \leqslant 7\)
B2
ft their graph if incomplete eg no graph drawn for \(-2 \leqslant x \lt 0\) but otherwise correct and answer \(-5 \leqslant \mathrm{f}(x) \leqslant 4\)
B2ft
ft their graph if drawn for \(x\) values beyond [\(-2\), 5] eg1 straight line from (\(-3\), 8) to (6, \(-1\)) and answer \(-1 \leqslant y \leqslant 8\) eg2 straight line from (\(-3\), 8) to (6, \(-1\)) and answer \(\mathrm{f}(x) \leqslant 8\)
B2ft B1ft
Straight line from (\(-2\), 9) to (6, \(-7\)) and answer \(-7 \leqslant y \leqslant 9\)
B2ft
Straight line from (0, 9) to (5, \(-4\)) and answer \(-4 \leqslant \mathrm{f}(x) \leqslant 9\)
B2ft
B2ft (or B1ft) can be awarded for a range beyond [\(-7\), 9] if it is clear from working (eg a table of values) where the answer is from
\(-5\) to 7 inclusive is B2 whereas \(-5\) to 7 is B1
B1 for a correct inequality embedded eg1 \(-5 \lt \mathrm{f}(x) \leqslant 7\) eg2 \(-5 \leqslant \mathrm{f}(x) \leqslant 0\) eg3 \(-2 \leqslant y \leqslant 7\)
B1 B1 B1
For B1 ignore incorrect notation if only \(-5\) and 7 chosen eg1 \(-5 \leqslant x \leqslant 7\) eg2 \(-5 \lt x \leqslant 7\) eg3 \(-5 \geqslant \mathrm{f}(x) \geqslant 7\) eg4 \(-5\), 7
B1 B1 B1 B1
{\(-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7\)}
B0
Working out a statistical range eg \(-5\) to \(7 = 12\)
19 The continuous curve \(\ y = \mathrm{g}(x)\ \) has exactly two stationary points.
The stationary points are
a point of inflection at \(P(1, 2)\)
a minimum point at \(Q(a, b)\) where \(a \gt 1\) and \(b \lt 0\)
On the axes below, sketch the curve.
Label points \(P\) and \(Q\) on your sketch. [3 marks]
Mark scheme
Answer
Mark
Comments
Continuous curve with point of inflection, labelled \(P\) or (1, 2), in first quadrant and minimum point, labelled \(Q\) or \((a, b)\), in fourth quadrant, with \(x\)-coordinate of Q > \(x\)-coordinate of \(P\) eg
B3
For B3, allow the labelling of one coordinate as sufficient for each point B2 As B3 but not sufficiently labelled B1 Curve with point of inflection, labelled \(P\) or (1, 2), in first quadrant or curve with minimum point, labelled \(Q\) or \((a, b)\), in fourth quadrant
For B1, allow labelling using one coordinate as sufficient
SC2 As B3 but \(x\)-coordinate of Q < \(x\)-coordinate of \(P\) eg
Additional guidance
For B3, curve does not have to cross the \(x\)-axis after \(Q\) and does not have to cross the \(y\)-axis before \(P\)
For a stationary point, curve must not stop at the point
At \(P\), the curve must change from concave upward to concave downward for B3 or B2 or vice-versa for B1 or SC2
Note that other non-stationary points of inflection may also be seen (up to B3 possible)
Curve may have horizontal asymptotes as \(x \to \pm\infty\) (up to B3 possible)
Mark intention for stationary points, positioning of labels and smoothness of curve
More than 1 stationary point of inflection and/or more than 1 minimum point and/or maximum point(s) can score a maximum of B1
Labelling using a coordinate or coordinates may be seen by labelling on an axis or on axes (axes may also show other numbers)
13 Here is a sketch of \(\quad y = a + bx - 2x^2 \quad\) where \(a\) and \(b\) are constants.
The graph intersects the \(x\)-axis at \((-1, 0)\) and \(\left(\dfrac{7}{2}, 0\right)\) and the \(y\)-axis at point \(P\).
Not drawn accurately
Work out the coordinates of point \(P\).
You must show your working. [4 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1
\(a - b - 2 = 0\) or \(a + \dfrac{7}{2}b - \dfrac{49}{2} = 0\)
M1
oe equation Allow an unsimplified equation eg \(a + b \times (-1) - 2(-1)^2 = 0\) Missing brackets can be recovered
\(a - b - 2 = 0\) and \(a + \dfrac{7}{2}b - \dfrac{49}{2} = 0\)
M1
oe two equations Allow unsimplified equations
\(\dfrac{7}{2}a + a = 7 + \dfrac{49}{2}\) or \(\dfrac{7}{2}b - -b = \dfrac{49}{2} - 2\) and \(a\) − their \(b - 2 = 0\)
M1dep
oe dep on first M1 Correct method to form an equation in \(a\) or correct method to form an equation in \(b\) and substitutes to form an equation in \(a\) Their two equations must both contain \(a\) and \(b\)
\((0, 7)\)
A1
SC4 Answer \((0, 7)\) from \((2x - 7)(1 + x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)
Alternative method 2
\((7 - 2x)\) or \((1 + x)\) or \((2x - 7)\) or \((-1 - x)\) or \((x - 3.5)\) or \((3.5 - x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)
M1
oe Brackets not needed
\((7 - 2x)(1 + x)\) or \((2x - 7)(-1 - x)\) or \(7 + 7x - 2x - 2x^2\) or \(7 + 5x - 2x^2\)
M1
oe eg \(-(2x^2 - 5x - 7)\) \(y =\) not needed Expansion not needed
Substitutes \(x = 0\) in their quadratic or selects the constant term from their quadratic
M1dep
dep on first M1 Expansion not needed for their quadratic but must be correct if attempted May be implied by the final answer
\((0, 7)\)
A1
SC4 Answer \((0, 7)\) from \((2x - 7)(1 + x)\) or \(2x^2 + 2x - 7x - 7\) or \(2x^2 - 5x - 7\)
2 The function \(\quad \mathrm{f}(x) = x^2 - 4x + 1 \quad\) has domain \(\ -1 \leqslant x \leqslant 5\)
Here is the graph of \(\quad y = \mathrm{f}(x)\)
(a) Write down the equation of the line of symmetry of the graph. [1 mark]
(b) Use the graph to work out the solutions of \(\quad x^2 - 4x + 1 = 5\)
Give your answers to 1 decimal place. [2 marks]
(c) Write down the range of \(\mathrm{f}(x)\) for domain \(\ -1 \leqslant x \leqslant 5\) [2 marks]
Mark scheme (a)
Answer
Mark
Comments
\(x = 2\)
B1
Additional guidance
\(2 = x\)
B1
\(y = 2\)
B0
2
B0
Mark scheme (b)
Answer
Mark
Comments
\(-0.8\) and 4.8 with no other answers
B2
Both correct in either order B1 One correct and one incorrect or missing
Additional guidance
The word ‘and’ is not needed If their answer has both \(-0.8\) and 4.8 with no other solutions award B2 eg 4.8, \(-0.8\) with no other solutions
B2
Mark scheme (c)
Answer
Mark
Comments
\(-3 \leqslant \mathrm{f}(x) \leqslant 6\) or \(6 \geqslant \mathrm{f}(x) \geqslant -3\) or [\(-3\), 6]
B2
oe
B1 \(\mathrm{f}(x) \geqslant -3\) or \(\mathrm{f}(x) \leqslant 6\) on their own or embedded within an interval for \(\mathrm{f}(x)\) or only \(-3\) and 6 chosen
Additional guidance
Allow as two inequalities \(\mathrm{f}(x) \geqslant -3 \quad \mathrm{f}(x) \leqslant 6\)
B2
\(-3\) to 6 inclusive
B2
\(\mathrm{f}(x)\) may be replaced by f or \(y\) or \(x^2 - 4x + 1\) for B2 and B1
B1 may be seen with an incorrect inequality eg1 \(-3 \lt \mathrm{f}(x) \leqslant 6\) eg2 \(-3 \leqslant \mathrm{f}(x) \lt 6\) eg3 \(-3 \leqslant \mathrm{f}(x) \leqslant 5\)
B1 B1 B1
For B1 ignore incorrect notation if only \(-3\) and 6 chosen eg1 \(-3 \leqslant x \leqslant 6\) eg2 \(-3 \lt x \leqslant 6\) eg3 \(-3\) to 6 eg4 \(-3\) to \(6 = 9\) (working out a statistical range) eg5 \(-3 \geqslant \mathrm{f}(x) \geqslant 6\) eg6 \(-3\), 6
