IB DP Mathematical Studies 1.0 Basic manipulation of simple algebraic expressions, including factorization and expansion Paper 1

 

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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 \({x^2} – 3x – 10\).

[2]
a.

A function is defined as \(f(x) = 1 + {x^3}\) for \(x \in \mathbb{Z}{\text{, }} {- 3} \leqslant x \leqslant 3\).

(i)     List the elements of the domain of \(f(x)\).

(ii)    Write down the range of \(f(x)\).

[4]
b.
Answer/Explanation

Markscheme

\((x – 5)(x + 2)\)     (A1)(A1)     (C2)

Note: Award (A1) for \((x + 5)(x – 2)\), (A0) otherwise. If equation is equated to zero and solved by factorizing award (A1) for both correct factors, followed by (A0).

[2 marks]

a.

(i)     \( – 3\), \( – 2\), \( – 1\), \(0\), \(1\), \(2\), \(3\)     (A1)(A1)     (C2)


Note:
Award (A2) for all correct answers seen and no others. Award (A1) for 3 correct answers seen.

(ii)    \( – 26\), \( – 7\), 0,1, 2, 9, 28     (A1)(A1)     (C2)


Note:
Award (A2) for all correct answers seen and no others. Award (A1) for 3 correct answers seen. If domain and range are interchanged award (A0) for (b)(i) and (A1)(ft)(A1)(ft) for (b)(ii).

[4 marks]

b.

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

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

\(10 000\) people attended a sports match. Let \(x\) be the number of adults attending the sports match and \(y\) be the number of children attending the sports match.

Write down an equation in \(x\) and \(y\) .

[1]
a.

The cost of an adult ticket was \(12\) Australian dollars (AUD). The cost of a child ticket was \(5\) Australian dollars (AUD).

Find the total cost for a family of 2 adults and 3 children.

[2]
b.

The total cost of tickets sold for the sports match was \(108800{\text{ AUD}}\).

Write down a second equation in \(x\) and \(y\) .

[1]
c.

Write down the value of \(x\) and the value of \(y\) .

[2]
d.
Answer/Explanation

Markscheme

\(x + y = 10000\)     (A1)     (C1)

[1 mark]

a.

\(2 \times 12 + 3 \times 5\)     (M1)

\(39{\text{ }}(39.0{\text{, }}39.00)\) (AUD)     (A1)     (C2)

[2 marks]

b.

\(12x + 5y = 108800\)     (A1)     (C1)

[1 mark]

c.

\(x = 8400\), \(y = 1600\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Follow through from their equations. If \(x\) and \(y\) are both incorrect then award (M1) for attempting to solve simultaneous equations.

[2 marks]

d.

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

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

Expand the expression \(x(2{x^3} – 1)\).

[2]
a.

Differentiate \(f(x) = x(2{x^3} – 1)\).

[2]
b.

Find the \(x\)-coordinate of the local minimum of the curve \(y = f(x)\).

[2]
c.
Answer/Explanation

Markscheme

\(2{x^4} – x\)     (A1)(A1)     (C2)

Note: Award (A1) for \(2{x^4}\), (A1) for \( – x\).

[2 marks]

a.

\(8{x^3} – 1\)     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) for \(8{x^3}\), (A1)(ft) for \(–1\). Follow through from their part (a).

     Award at most (A1)(A0) if extra terms are seen.

[2 marks]

b.

\(8{x^3} – 1 = 0\)     (M1)

Note: Award (M1) for equating their part (b) to zero.

\((x = )\frac{1}{2}{\text{ (0.5)}}\)     (A1)(ft)     (C2)

Notes: Follow through from part (b).

     \(0.499\) is the answer from the use of trace on the GDC; award (A0)(A0).

     For an answer of \((0.5, –0.375)\), award (M1)(A0).

[2 marks]

c.
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