Question
The length of one side of a rectangle is 2 cm longer than its width.
If the smaller side is x cm, find the perimeter of the rectangle in terms of x.[1]
The length of one side of a rectangle is 2 cm longer than its width.
The perimeter of a square is equal to the perimeter of the rectangle in part (a).
Determine the length of each side of the square in terms of x.[1]
The length of one side of a rectangle is 2 cm longer than its width.
The perimeter of a square is equal to the perimeter of the rectangle in part (a).
The sum of the areas of the rectangle and the square is \(2x^2 + 4x +1\) (cm2).
(i) Given that this sum is 49 cm2, find x.
(ii) Find the area of the square.[4]
Answer/Explanation
Markscheme
Unit penalty (UP) is applicable where indicated in the left hand column.
(UP) \({\text{P (rectangle)}} = 2x + 2(x + 2) = 4x + 4{\text{ cm}}\) (A1) (C1)
(UP) Simplification not required[1 mark]
Unit penalty (UP) is applicable where indicated in the left hand column.
(UP) Side of square = (4x + 4)/4 = x + 1 cm (A1)(ft) (C1)[1 mark]
(i) \(2x^2 + 4x + 1 = 49\) or equivalent (M1)
\((x + 6)(x – 4) = 0\)
\(x = – 6\) and \(4\) (A1)
Note: award (A1) for the values or for correct factors
Choose \(x = 4\) (A1)(ft)
Award (A1)(ft) for choosing positive value. (C3)
(ii) \({\text{Area of square}} = 5 \times 5 = 25{\text{ c}}{{\text{m}}^2}\) (A1)(ft)
Note: Follow through from both (b) and (c)(i). (C1)[4 marks]
Question
Solve the following equation for x
\(3(2x +1) − 2(3 − x)=13\).[2]
Factorize the expression \(x^2 + 2x − 3\).[2]
Find the positive solution of the equation
\(x^2 + 2x − 6 = 0\).[2]
Answer/Explanation
Markscheme
\(6x+ 3 – 6 + 2x = 13\) (M1)
\(8x = 16\)
\(x = 2\) (A1) (C2)[2 marks]
\((x + 3) (x – 1)\) (A1)(A1) (C2)[2 marks]
\(x = 1.64575…\)
\(x = 1.65\) (A2)
If formula is used award (M1)(A1) for correct solution. If graph is sketched award (M1)(A1) for correct solution. (C2)[2 marks]
Question
Factorise the expression \({x^2} – kx\) .[1]
Hence solve the equation \({x^2} – kx = 0\) .[1]
The diagram below shows the graph of the function \(f(x) = {x^2} – kx\) for a particular value of \(k\).
Write down the value of \(k\) for this function.[1]
The diagram below shows the graph of the function \(f(x) = {x^2} – kx\) for a particular value of \(k\).
Find the minimum value of the function \(y = f(x)\) .[3]
Answer/Explanation
Markscheme
\(x(x – k)\) (A1) (C1)[1 mark]
\(x = 0\) or \(x = k\) (A1) (C1)
Note: Both correct answers only.[1 mark]
\(k = 3\) (A1) (C1)[1 mark]
\({\text{Vertex at }}x = \frac{{ – ( – 3)}}{{2(1)}}\) (M1)
Note: (M1) for correct substitution in formula.
\(x = 1.5\) (A1)(ft)
\(y = – 2.25\) (A1)(ft)
OR
\(f'(x) = 2x – 3\) (M1)
Note: (M1) for correct differentiation.
\(x = 1.5\) (A1)(ft)
\(y = – 2.25\) (A1)(ft)
OR
for finding the midpoint of their 0 and 3 (M1)
\(x = 1.5\) (A1)(ft)
\(y = – 2.25\) (A1)(ft)
Note: If final answer is given as \((1.5{\text{, }}{- 2.25})\) award a maximum of (M1)(A1)(A0)[3 marks]
Question
Let \(f (x) = x^2 – 6x + 8\).
Factorise \(x^2 – 6x + 8\).[2]
Hence, or otherwise, solve the equation \(x^2 – 6x + 8 = 0\).[2]
Let \(g(x) = x + 3\).
Write down the solutions to the equation \(f (x) = g(x)\).[2]
Answer/Explanation
Markscheme
\( (x – 2)(x – 4)\) (A1)(A1) (C2)[2 marks]
x = 2, x = 4 (A1)(ft)(A1)(ft) (C2)[2 marks]
x = 0.807, x = 6.19 (A1)(A1) (C2)
Note: Award maximum of (A0)(A1) if coordinate pairs given.
OR
(M1) for an attempt to solve \(x^2 – 7x + 5 = 0\) via formula with correct values substituted. (M1)
\(x = \frac{{7 \pm \sqrt {29} }}{2}\) (A1) (C2)[2 marks]
Question
The length of a square garden is (x + 1) m. In one of the corners a square of 1 m length is used only for grass. The rest of the garden is only for planting roses and is shaded in the diagram below.
The area of the shaded region is A .
Write down an expression for A in terms of x.[1]
Find the value of x given that A = 109.25 m2.[3]
The owner of the garden puts a fence around the shaded region. Find the length of this fence.[2]
Answer/Explanation
Markscheme
(x + 1)2 – 1 or x2 + 2x (A1) (C1)[1 mark]
(x + 1)2 – 1 = 109.25 (M1)
x2 + 2x – 109.25 = 0 (M1)
Notes: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for correctly removing the brackets.
OR
(x + 1)2 – 1 = 109.25 (M1)
x + 1 = \(\sqrt {110.25} \) (M1)
Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for taking the square root of both sides.
OR
(x + 1)2 – 10.52 = 0 (M1)
(x – 9.5) (x + 11.5) = 0 (M1)
Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for factorised left side of the equation.
x = 9.5 (A1)(ft) (C3)
Note: Follow through from their expression in part (a).
The last mark is lost if x is non positive.
If the follow through equation is not quadratic award at most (M1)(M0)(A1)(ft).[3 marks]
4 × (9.5 + 1) = 42 m (M1)(A1)(ft) (C2)
Notes: Award (M1) for correct method for finding the length of the fence. Accept equivalent methods.[2 marks]
Question
In the diagram, \({\text{B}}\hat {\text{A}}{\text{C}} = {90^ \circ }\) . The length of the three sides are \(x{\text{ cm}}\), \((x + 7){\text{ cm}}\) and \((x + 8){\text{ cm}}\).
Write down and simplify a quadratic equation in \(x\) which links the three sides of the triangle.[3]
Solve the quadratic equation found in part (a).[2]
Write down the value of the perimeter of the triangle.[1]
Answer/Explanation
Markscheme
\({(x + 8)^2} = {(x + 7)^2} + {x^2}\) (A1)
Note: Award (A1) for a correct equation.
\({x^2} + 16x + 64 = {x^2} + 14x + 49 + {x^2}\) (A1)
Note: Award (A1) for correctly removed parentheses.
\({x^2} – 2x – 15 = 0\) (A1) (C3)
Note: Accept any equivalent form.[3 marks]
\(x = 5\), \(x = – 3\) (A1)(ft)(A1)(ft) (C2)
Notes: Accept (A1)(ft) only from the candidate’s quadratic equation.[2 marks]
\(30{\text{ cm}}\) (A1)(ft) (C1)
Note: Follow through from a positive answer found in part (b).[1 mark]
Question
In the diagram, triangle ABC is isosceles. AB = AC and angle ACB is 32°. The length of side AC is x cm.
Write down the size of angle CBA.
Write down the size of angle CAB.[1]
The area of triangle ABC is 360 cm2. Calculate the length of side AC. Express your answer in millimetres.[4]
Answer/Explanation
Markscheme
32° (A1) (C1)[1 mark]
116° (A1) (C1)[1 mark]
\(360 = \frac{1}{2} \times {x^2} \times \sin 116^\circ \) (M1)(A1)(ft)
Notes: Award (M1) for substitution into correct formula with 360 seen, (A1)(ft) for correct substitution, follow through from their answer to part (b).
x = 28.3 (cm) (A1)(ft)
x = 283 (mm) (A1)(ft) (C4)
Notes: The final (A1)(ft) is for their cm answer converted to mm. If their incorrect cm answer is seen the final (A1)(ft) can be awarded for correct conversion to mm.[4 marks]
Question
f (x) = 5x3 − 4x2 + x
Find f‘(x).[3]
Find using your answer to part (a) the x-coordinate of
(i) the local maximum point;
(ii) the local minimum point.[3]
Answer/Explanation
Markscheme
15x2 – 8x + 1 (A1)(A1)(A1) (C3)
Note: Award (A1) for each correct term.[3 marks]
15x2 – 8x +1 = 0 (A1)(ft)
Note: Award (A1)(ft) for setting their derivative to zero.
(i) \((x =)\frac{1}{5}(0.2)\) (A1)(ft)
(ii) \((x =)\frac{1}{3}(0.333)\) (A1)(ft) (C3)
Notes: Follow through from their answer to part (a).[3 marks]
Question
The graph of the quadratic function \(f (x) = c + bx − x^2\) intersects the y-axis at point A(0, 5) and has its vertex at point B(2, 9).
Write down the value of c.[1]
Find the value of b.[2]
Find the x-intercepts of the graph of f .[2]
Write down \(f (x)\) in the form \(f (x) = −(x − p) (x + q)\).[1]
Answer/Explanation
Markscheme
5 (A1) (C1)
\(\frac{{ – b}}{{2( – 1)}} = 2\) (M1)
Note: Award (M1) for correct substitution in axis of symmetry formula.
OR
\(y = 5 + bx – x^2\)
\(9 = 5 + b (2) – (2)^2\) (M1)
Note: Award (M1) for correct substitution of 9 and 2 into their quadratic equation.
\((b =) 4\) (A1)(ft) (C2)
Note: Follow through from part (a).
5, −1 (A1)(ft)(A1)(ft) (C2)
Notes: Follow through from parts (a) and (b), irrespective of working shown.
\(f (x) = -(x – 5)(x + 1)\) (A1)(ft) (C1)
Notes: Follow through from part (c).
Question
A quadratic function \(f:x \mapsto a{x^2} + b\), where \(a\) and \(b \in \mathbb{R}\) and \(x \geqslant 0\), is represented by the mapping diagram.
Using the mapping diagram, write down two equations in terms of \(a\) and \(b\).[2]
Solve the equations to find the value of
(i) \(a\);
(ii) \(b\).[2]
Find the value of \(c\).[2]
Answer/Explanation
Markscheme
\(a{(1)^2} + b = – 9\) (A1)
\(a{(3)^2} + b = 119\) (A1) (C2)
Note: Accept equivalent forms of the equations.[2 marks]
(i) \(a = 16\) (A1)(ft)
(ii) \(b = – 25\) (A1)(ft) (C2)
Note: Follow through from part (a) irrespective of whether working is seen.
If working is seen follow through from part (i) to part (ii).[2 marks]
\(16{c^2} – 25 = 171\) (M1)
Note: Award (M1) for correct quadratic with their \(a\) and \(b\) substituted.
\(c = 3.5\) (A1)(ft) (C2)
Note: Accept \(x\) instead of \(c\).
Follow through from part (b).
Award (A1) only, for an answer of \( \pm 3.5\) with or without working.[2 marks]
Question
The surface of a red carpet is shown below. The dimensions of the carpet are in metres.
Write down an expression for the area, \(A\), in \({{\text{m}}^2}\), of the carpet.[1]
The area of the carpet is \({\text{10 }}{{\text{m}}^2}\).
Calculate the value of \(x\).[3]
The area of the carpet is \({\text{10 }}{{\text{m}}^2}\).
Hence, write down the value of the length and of the width of the carpet, in metres.[2]
Answer/Explanation
Markscheme
\(2x(x – 4)\) or \(2{x^2} – 8x\) (A1) (C1)
Note: Award (A0) for \(x – 4 \times 2x\).[1 mark]
\(2x(x – 4) = 10\) (M1)
Note: Award (M1) for equating their answer in part (a) to \(10\).
\({x^2} – 4x – 5 = 0\) (M1)
OR
Sketch of \(y = 2{x^2} – 8x\) and \(y = 10\) (M1)
OR
Using GDC solver \(x = 5\) and \(x = – 1\) (M1)
OR
\(2(x + 1)(x – 5)\) (M1)
\(x = 5{\text{ (m)}}\) (A1)(ft) (C3)
Notes: Follow through from their answer to part (a).
Award at most (M1)(M1)(A0) if both \(5\) and \(-1\) are given as final answer.
Final (A1)(ft) is awarded for choosing only the positive solution(s).[3 marks]
\(2 \times 5 = 10{\text{ (m)}}\) (A1)(ft)
\(5 – 4 = 1{\text{ (m)}}\) (A1)(ft) (C2)
Note: Follow through from their answer to part (b).
Do not accept negative answers.[2 marks]
Question
A building company has many rectangular construction sites, of varying widths, along a road.
The area, \(A\), of each site is given by the function
\[A(x) = x(200 – x)\]
where \(x\) is the width of the site in metres and \(20 \leqslant x \leqslant 180\).
Site S has a width of \(20\) m. Write down the area of S.[1]
Site T has the same area as site S, but a different width. Find the width of T.[2]
When the width of the construction site is \(b\) metres, the site has a maximum area.
(i) Write down the value of \(b\).
(ii) Write down the maximum area.[2]
The range of \(A(x)\) is \(m \leqslant A(x) \leqslant n\).
Hence write down the value of \(m\) and of \(n\).[1]
Answer/Explanation
Markscheme
\(3600{\text{ (}}{{\text{m}}^2})\) (A1)(C1)
\(x(200 – x) = 3600\) (M1)
Note: Award (M1) for setting up an equation, equating to their \(3600\).
\(180{\text{ (m)}}\) (A1)(ft) (C2)
Note: Follow through from their answer to part (a).
(i) \(100{\text{ (m)}}\) (A1) (C1)
(ii) \(10\,000{\text{ (}}{{\text{m}}^2})\) (A1)(ft)(C1)
Note: Follow through from their answer to part (c)(i).
\(m = 3600\;\;\;\)and\(\;\;\;n = 10\,000\) (A1)(ft) (C1)
Notes: Follow through from part (a) and part (c)(ii), but only if their \(m\) is less than their \(n\). Accept the answer \(3600 \leqslant A \leqslant 10\,000\).