IB DP Mathematical Studies 1.0 Solving linear equations in one variable 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

Water has a lower boiling point at higher altitudes. The relationship between the boiling point of water (T) and the height above sea level (h) can be described by the model \(T = -0.0034h +100\) where T is measured in degrees Celsius (°C) and h is measured in metres from sea level.

Write down the boiling point of water at sea level.

[1]
a.

Use the model to calculate the boiling point of water at a height of 1.37 km above sea level.

[3]
b.

Water boils at the top of Mt. Everest at 70 °C.

Use the model to calculate the height above sea level of Mt. Everest.

[2]
c.
Answer/Explanation

Markscheme

100 °C     (A1)     (C1)

[1 mark]

a.

\(T = -0.0034 \times 1370 + 100\)     (A1)(M1)

Note: Award (A1) for 1370 seen, (M1) for substitution of their h into the equation.

95.3 °C (95.342)     (A1)     (C3)

Notes: If their h is incorrect award at most (A0)(M1)(A0). If their h = 1.37 award at most (A0)(M1)(A1)(ft).

[3 marks]

b.

\(70 = -0.0034h + 100\)     (M1)

Note: Award (M1) for correctly substituted equation.

h = 8820 m (8823.52…)     (A1)     (C2)

Note: The answer is 8820 m (or 8.82 km.) units are required.

[2 marks]

c.

Question

In an arithmetic sequence, the fifth term, u5, is greater than the first term, u1. The difference between these terms is 36.

Find the common difference, d.

[2]
a.

The tenth term of the sequence is double the seventh term.

(i) Write down an equation in u1 and d to show this information.

(ii) Find u1.

[4]
b.
Answer/Explanation

Markscheme

\({u_1} + 4d – {u_1} = 36\)     (M1)

 

Note: Accept equivalent forms including the use of a instead of \({u_1}\).

\((d =) 9\)     (A1)     (C2)

a.

(i) \({u_{10}} = 2{u_7}\)     (M1)

Note: Award (M1) for correct use of 2 (may be implied).

\({u_1} + 9d = 2[{u_1} + 6d]\)     (A1)

Notes: Accept equivalent forms. Award (M1)(A0) for \(a + 9d = 2[a + 6d]\).

(ii) \({u_1} + 81 = 2{u_1} + 108\)     (M1)

\(({u_1} =) – 27\)     (A1)(ft)     (C4)

Notes: Follow through from their d found in part (a) and equation in (b)(i). Do not penalize further use of a instead of \({u_1}\).

b.

Question

Consider the function \(f (x) = ax^3 − 3x + 5\), where \(a \ne 0\).

Find \(f ‘ (x) \).

[2]
a.

Write down the value of \(f ′(0)\).

[1]
b.

The function has a local maximum at x = −2.

Calculate the value of a.

[3]
c.
Answer/Explanation

Markscheme

\( f ‘(x) = 3ax^2 – 3\)     (A1)(A1)     (C2)

Note: Award a maximum of (A1)(A0) if any extra terms are seen.

a.

−3     (A1)(ft)     (C1)

Note: Follow through from their part (a).

b.

\(f ‘(x) = 0\)     (M1)

Note: This may be implied from line below.

\(3a(-2)^2 – 3 = 0\)     (M1)

\((a =) \frac{1}{4}\)     (A1)(ft)     (C3)

Note: Follow through from their part (a).

c.

Question

The diagram below represents a rectangular flag with dimensions 150 cm by 92 cm. The flag is divided into three regions A, B and C.

Write down the total area of the flag.

[1]
a.

Write down the value of y.

[1]
b.

The areas of regions A, B, and C are equal.

Write down the area of region A.

[1]
c.

Using your answers to parts (b) and (c), find the value of x.

[3]
d.
Answer/Explanation

Markscheme

Units are required in this question for full marks to be awarded.

13800 cm2     (A1)     (C1)

a.

75     (A1)     (C1)

b.

Units are required in this question for full marks to be awarded.

4600 cm2     (A1)(ft)     (C1)

Notes: Units are required unless already penalized in part (a). Follow through from their part (a).

c.

\(0.5(x + 92) \times 75 = 4600\)     (M1)(A1)(ft)

OR

\(0.5 \times 150 \times (92 – x) = 4600\)     (M1)(A1)(ft)

Note: Award (M1) for substitution into area formula, (A1)(ft) for their correct substitution.

(= 30.7 (cm)(30.6666…(cm))     (A1)(ft)     (C3)

Note: Follow through from their parts (b) and (c).

d.

Question

A liquid is heated so that after \(20\) seconds of heating its temperature, \(T\) , is \({25^ \circ }{\text{C}}\) and after \(50\) seconds of heating its temperature is \({37^ \circ }{\text{C}}\) .

The temperature of the liquid at time \(t\) can be modelled by \(T = at + b\) , where \(t\) is the time in seconds after the start of heating.

Using this model one equation that can be formed is \(20a + b = 25\) .

Using the model, write down a second equation in \(a\) and \(b\) .

[2]
a.

Using your graphic display calculator or otherwise, find the value of \(a\) and of \(b\) .

[2]
b.

Use the model to predict the temperature of the liquid \(60{\text{ seconds}}\) after the start of heating.

[2]
c.
Answer/Explanation

Markscheme

\(50a + b = 37\) (A1)(A1)     (C2)

Note: Award (A1) for \(50a + b\) , (A1) for \(= 37\) .

a.

\(a = 0.4\), \(b = 17\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (M1) for attempt to solve their equations if this is done analytically. If the GDC is used, award (ft) even if no working seen.

b.

\(T = 0.4(60) + 17\)     (M1)

Note: Award (M1) for correct substitution of their values and \(60\) into equation for \(T\).

\(T = 41{\text{ }}{{\text{(}}^ \circ }{\text{C}})\)     (A1)(ft)     (C2)

Note: Follow through from their part (b).

c.
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