IB DP Mathematical Studies 1.6 Use of a GDC to solve quadratic equations Paper 1

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]

a.

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]

b.

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]

c.
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]

a.

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]

b.

(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]

c.

Question

Solve the following equation for x

\(3(2x +1) − 2(3 − x)=13\).[2]

a.

Factorize the expression \(x^2 + 2x − 3\).[2]

b.

Find the positive solution of the equation

\(x^2 + 2x − 6 = 0\).[2]

c.
Answer/Explanation

Markscheme

\(6x+ 3 – 6 + 2x = 13\)     (M1)

\(8x = 16\)

\(x = 2\)     (A1)     (C2)[2 marks]

a.

\((x + 3) (x – 1)\)     (A1)(A1)     (C2)[2 marks]

b.

\(x = 1.64575…\)

\(x = 1.65\)     (A2)

If formula is used award (M1)(A1) for correct solution. If graph is sketched award (M1)(A1) for correct solution.     (C2)[2 marks]

c.

Question

Factorise the expression \({x^2} – kx\) .[1]

a.

Hence solve the equation \({x^2} – kx = 0\) .[1]

b.

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]

c.

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]

d.
Answer/Explanation

Markscheme

\(x(x – k)\)     (A1)     (C1)[1 mark]

a.

\(x = 0\) or \(x = k\)     (A1)     (C1)

Note: Both correct answers only.[1 mark]

b.

\(k = 3\)     (A1)     (C1)[1 mark]

c.

\({\text{Vertex at }}x = \frac{{ – ( – 3)}}{{2(1)}}\)     (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]

d.

Question

Let \(f (x) = x^2 – 6x + 8\).

Factorise \(x^2 – 6x + 8\).[2]

a.

Hence, or otherwise, solve the equation \(x^2 – 6x + 8 = 0\).[2]

b.

Let \(g(x) = x + 3\).

Write down the solutions to the equation \(f (x) = g(x)\).[2]

c.
Answer/Explanation

Markscheme

\( (x – 2)(x – 4)\)     (A1)(A1)     (C2)[2 marks]

a.

x = 2, x = 4     (A1)(ft)(A1)(ft)     (C2)[2 marks]

b.

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]

c.

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]

a.

Find the value of x given that A = 109.25 m2.[3]

b.

The owner of the garden puts a fence around the shaded region. Find the length of this fence.[2]

c.
Answer/Explanation

Markscheme

(x + 1)2 – 1  or  x2 + 2x     (A1)     (C1)[1 mark]

a.

(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]

b.

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]

c.

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]

a.

Solve the quadratic equation found in part (a).[2]

b.

Write down the value of the perimeter of the triangle.[1]

c.
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]

a.

\(x = 5\), \(x = – 3\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept (A1)(ft) only from the candidate’s quadratic equation.[2 marks]

b.

\(30{\text{ cm}}\)     (A1)(ft)     (C1)

Note: Follow through from a positive answer found in part (b).[1 mark]

c.

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.

[1]
a.

Write down the size of angle CAB.[1]

b.

The area of triangle ABC is 360 cm2. Calculate the length of side AC. Express your answer in millimetres.[4]

c.
Answer/Explanation

Markscheme

32°     (A1)     (C1)[1 mark]

a.

116°     (A1)     (C1)[1 mark]

b.

\(360 = \frac{1}{2} \times {x^2} \times \sin 116^\circ \)     (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]

c.

Question

f (x) = 5x3 − 4x2 + x

Find f‘(x).[3]

a.

Find using your answer to part (a) the x-coordinate of

(i) the local maximum point;

(ii) the local minimum point.[3]

b.
Answer/Explanation

Markscheme

15x2 – 8x + 1     (A1)(A1)(A1)     (C3)

Note: Award (A1) for each correct term.[3 marks]

a.

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]

b.

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]

a.

Find the value of b.[2]

b.

Find the x-intercepts of the graph of f .[2]

c.

Write down \(f (x)\) in the form \(f (x) = −(x − p) (x + q)\).[1]

d.
Answer/Explanation

Markscheme

5     (A1)     (C1)

a.

\(\frac{{ – b}}{{2( – 1)}} = 2\)     (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).

b.

5, −1     (A1)(ft)(A1)(ft)     (C2)

Notes: Follow through from parts (a) and (b), irrespective of working shown.

c.

\(f (x) = -(x – 5)(x + 1)\)     (A1)(ft)     (C1)

Notes: Follow through from part (c).

d.

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]

a.

Solve the equations to find the value of

(i)     \(a\);

(ii)     \(b\).[2]

b.

Find the value of \(c\).[2]

c.
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]

a.

(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]

b.

\(16{c^2} – 25 = 171\)     (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]

c.

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]

a.

The area of the carpet is \({\text{10 }}{{\text{m}}^2}\).

Calculate the value of \(x\).[3]

b.

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]

c.
Answer/Explanation

Markscheme

\(2x(x – 4)\)   or   \(2{x^2} – 8x\)     (A1)     (C1)

Note: Award (A0) for \(x – 4 \times 2x\).[1 mark]

a.

\(2x(x – 4) = 10\)     (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]

b.

\(2 \times 5 = 10{\text{ (m)}}\)     (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]

c.

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]

a.

Site T has the same area as site S, but a different width. Find the width of T.[2]

b.

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]

c.

The range of \(A(x)\) is \(m \leqslant A(x) \leqslant n\).

Hence write down the value of \(m\) and of \(n\).[1]

d.
Answer/Explanation

Markscheme

\(3600{\text{ (}}{{\text{m}}^2})\)     (A1)(C1)

a.

\(x(200 – x) = 3600\)     (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).

b.

(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).

c.

\(m = 3600\;\;\;\)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 \(3600 \leqslant A \leqslant 10\,000\).

d.
Scroll to Top