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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ1 paper, which your school or the IB can provide.

Practice mode

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Section A · Short response

Questions 1–9 · 56 marks

Question 1

5 marks

A definite integral with a square root

Number and algebra

5 marks

Split the integrand.
Integrate term by term.
Evaluate between 1 and 9.

\frac{3 x - 5}{x} = 3 - 5 x^{- 1/2} \int (3 - 5 x^{- 1/2}) d x = 3 x - 10 x^{1/2} + c [3 x - 10 x]_{1}^{9} = (27 - 30) - (3 - 10) = - 3 + 7 = 4

Question 2

7 marks

A box plot, outliers and two regression lines

Statistics and probability

(a)

3 marks

The quartiles are $Q_{1} = 6$ and $Q_{3} = 10$ , so the IQR is 4. A value is an outlier above $Q_{3} + 1.5 \times IQR$ .
So the largest value that is not an outlier is $c = 16$ .

10 + 1.5 \times 4 = 16

(b)

4 marks

(i) To estimate the child's age from the adult's, use the regression line of $c$ on $a$ .

(ii) Both lines pass through $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Substitute $c = \frac{1}{2} a - 9$ into $a = \frac{7}{4} c + 20$ .

The mean age of the adults is 34 years.

c = \frac{1}{2} (42) - 9 = 12 years a = \frac{7}{4} (\frac{1}{2} a - 9) + 20 = \frac{7}{8} a + 4.25 \Rightarrow \frac{1}{8} a = 4.25 \Rightarrow a = 34

Question 3

7 marks

A composite of trig functions and an equation

Functions

(a)

2 marks

Replace $x$ by $2 x$ in $f$ .

(f \circ g) (x) = f (2 x) = 3 sin 2 x + cos 2 x

(b)

5 marks

Set $3 sin 2 x + cos 2 x = 2 cos 2 x$ , so $3 sin 2 x = cos 2 x$ .
For $0 \leq x \leq π$ we need $2 x$ in $[0, 2 π]$ : the tangent equals $\frac{1}{3}$ at $\frac{π}{6}$ and $\frac{7 π}{6}$ .

tan 2 x = \frac{1}{3} 2 x = \frac{π}{6}, \frac{7 π}{6} \Rightarrow x = \frac{π}{12}, \frac{7 π}{12}

Question 4

5 marks

A tangent parallel to a given line

Number and algebra

5 marks

Use the product rule on $y = (2 x - 1) e^{k x}$ .
At $x = 1$ the gradient is $2 e^{k} + k e^{k} = (k + 2) e^{k}$ . The line is $y = 5 e^{k} x$ , with gradient $5 e^{k}$ . Set them equal and divide by $e^{k}$ .

\frac{d y}{d x} = 2 e^{k x} + k (2 x - 1) e^{k x} (k + 2) e^{k} = 5 e^{k} \Rightarrow k + 2 = 5 \Rightarrow k = 3

Question 5

7 marks

Describing two transformations, then the smallest $y$ -intercept

Functions

(a)

2 marks

The graph of $g (x) = 4 sin (x - \frac{3 π}{2}) + 2.5 + q$ comes from $f$ by a translation of $\frac{3 π}{2}$ to the right (positive $x$ direction), and a translation of $q$ upwards (positive $y$ direction).

(b)

5 marks

Every value of $g$ is at least $7$ . The least value of the sine term is $- 4$ , so
The $y$ -intercept is $g (0)$ . Since $sin (- \frac{3 π}{2}) = 1$ :
The smallest value of $r$ comes from the smallest $q$ .

- 4 + 2.5 + q \geq 7 \Rightarrow q \geq 8.5 r = 4 (1) + 2.5 + q = 6.5 + q r_{m i n} = 6.5 + 8.5 = 15

Question 6

5 marks

The constant term of a binomial expansion

Number and algebra

5 marks

The general term of $(8 x^{3} - \frac{1}{2 x})^{n}$ is
The power of $x$ is $3 (n - r) - r = 3 n - 4 r$ . For a constant term this must be 0.
$r$ must be a whole number from 0 to $n$ , so $n$ must be a multiple of 4. The coefficient is then non-zero.

T_{r + 1} = (r n) (8 x^{3})^{n - r} (- \frac{1}{2 x})^{r} 3 n - 4 r = 0 \Rightarrow r = \frac{3 n}{4} n = 4, 8, 12, \dots (n a positive multiple of 4)

Question 7

8 marks

A density function with a square root: finding $k$ and the mean

Calculus

(a)

4 marks

The integral of $\frac{1}{4 - 3 x ^{2}}$ is $\frac{1}{3} arcsin \frac{3 x}{2}$ . Set the total area to 1.

\int_{0}^{1} \frac{k}{4 - 3 x ^{2}} d x = \frac{k}{3} arcsin \frac{3}{2} = \frac{k}{3} \cdot \frac{π}{3} = 1 k = \frac{3 3}{π}

(b)

4 marks

$E (X) = \int_{0}^{1} \frac{k x}{4 - 3 x ^{2}} d x$ . The numerator is nearly the derivative of the inside: $\frac{d}{d x} (4 - 3 x^{2}) = - 6 x$ .

E (X) = k [- \frac{1}{3} 4 - 3 x^{2}]_{0}^{1} = \frac{k}{3} (2 - 1) = \frac{k}{3} = \frac{3}{π}

Question 8

6 marks

Proof by contradiction: two odd numbers can't have squares summing to a multiple of 4

Number and algebra

6 marks

Assume that $a$ and $b$ are both odd, so $a = 2 m + 1$ and $b = 2 n + 1$ for some integers $m$ and $n$ .
This is a multiple of 4 plus 2, so it leaves remainder 2 when divided by 4. It is not divisible by 4. That contradicts the assumption that $a^{2} + b^{2}$ is exactly divisible by 4.
So $a$ and $b$ cannot both be odd.

a^{2} + b^{2} = (4 m^{2} + 4 m + 1) + (4 n^{2} + 4 n + 1) = 4 (m^{2} + m + n^{2} + n) + 2

Question 9

6 marks

Two complex numbers: their product and an argument condition

Number and algebra

(a)

3 marks

Multiply out and collect the real and imaginary parts.

z_{1} z_{2} = (1 + b i) ((1 - b^{2}) - 2 b i) = (1 - b^{2}) - 2 b i + b (1 - b^{2}) i - 2 b^{2} i^{2} z_{1} z_{2} = (1 + b^{2}) + i (- b - b^{3}) = (1 + b^{2}) (1 - b i)

(b)

3 marks

The argument is $arctan (\frac{- b ( 1 + b ^{2} )}{1 + b ^{2}}) = arctan (- b)$ , since $1 + b^{2} \neq = 0$ .

arctan (- b) = \frac{π}{4} \Rightarrow - b = 1 \Rightarrow b = - 1

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

18 marks

A series that is geometric, and then arithmetic

Number and algebra

(a)

6 marks

The terms are $ln x$ , $p ln x$ , $\frac{1}{3} ln x$ . (i) For a geometric series the ratios are equal.
(ii) The common ratio is $r = p = \pm \frac{1}{3}$ , so $∣ r ∣ < 1$ and the series converges.
(iii) For $p > 0$ , $r = \frac{1}{3}$ , and the sum to infinity is

\frac{p ln x}{ln x} = \frac{\frac{1}{3} ln x}{p ln x} \Rightarrow p = \frac{1}{3 p} \Rightarrow p^{2} = \frac{1}{3} \Rightarrow p = \pm \frac{1}{3} S_{\infty} = \frac{ln x}{1 - \frac{1}{3}} = 3 + 3 \Rightarrow ln x = (3 + 3) (1 - \frac{1}{3}) = 2 x = e^{2}

(b)

12 marks

(i) For an arithmetic series consecutive differences are equal.

(ii) The common difference is the first term minus the second, with the sign reversed.

(iii) The sum of the first $n$ terms is $S_{n} = \frac{n}{2} [2 ln x + (n - 1) d]$ , and it equals $ln (\frac{1}{x ^{3}}) = - 3 ln x$ .

p ln x - ln x = \frac{1}{3} ln x - p ln x \Rightarrow 2 p = \frac{4}{3} \Rightarrow p = \frac{2}{3} d = p ln x - ln x = - \frac{1}{3} ln x \frac{n}{2} [2 - \frac{n - 1}{3}] ln x = - 3 ln x \Rightarrow \frac{n ( 7 - n )}{6} = - 3 n^{2} - 7 n - 18 = 0 \Rightarrow (n - 9) (n + 2) = 0 \Rightarrow n = 9

Question 11

15 marks

Three planes with no common point, a line and a distance

Geometry and trigonometry

(a)

4 marks

The direction of the line where $Π_{1}$ and $Π_{2}$ meet is the cross product of their normals.
This direction is perpendicular to the normal of $Π_{3}$ , since $(- 9) (- 1) + 3 (- 5) + (- 2) (- 3) = 9 - 15 + 6 = 0$ . So the line is parallel to $Π_{3}$ .
A point on the line is $(1, - 2, 0)$ . It is not in $Π_{3}$ : $- 9 - 6 - 0 = - 15 \neq = 32$ . So the line never meets $Π_{3}$ , and the three planes have no common point.

2 - 1 1 \times 1 - 2 3 = - 1 - 5 - 3

(b)

5 marks

(i) For $P (1, - 2, 0)$ :

(ii) Use P and the direction vector from part (a).

Π_{1} : 2 + 2 + 0 = 4, Π_{2} : 1 + 4 + 0 = 5 r = 1 - 2 0 + λ 153

(c)

6 marks

The line is parallel to the plane, so every point of L is the same distance from $Π_{3}$ . Use the distance from P to $Π_{3}$ , whose normal is $(- 9, 3, - 2)$ with $∣ n ∣ = 94$ .

d = \frac{∣ - 9 ( 1 ) + 3 ( - 2 ) - 2 ( 0 ) - 32∣}{81 + 9 + 4} = \frac{47}{94} = \frac{94}{2} \approx 4.85

Question 12

21 marks

Maclaurin series of $e^{x} sin x$ and $e^{x} cos x$ , and a limit

Calculus

(a)

4 marks

Multiply the series for $e^{x}$ and $sin x$ and keep the terms up to $x^{3}$ .

(1 + x + \frac{x ^{2}}{2}) (x - \frac{x ^{3}}{6}) = x + x^{2} + \frac{x ^{3}}{2} - \frac{x ^{3}}{6} = x + x^{2} + \frac{1}{3} x^{3}

(b)

4 marks

Replace $x$ by $x^{2}$ in the series for $f$ , then integrate term by term.

\int_{0}^{1} e^{x^{2}} sin (x^{2}) d x \approx \int_{0}^{1} (x^{2} + x^{4} + \frac{1}{3} x^{6}) d x = \frac{1}{3} + \frac{1}{5} + \frac{1}{21} = \frac{61}{105}

(c)

5 marks

(i) Differentiate $g (x) = e^{x} cos x$ with the product rule twice.

Compare with $2 (g^{'} (x) - g (x))$ .

(ii) Differentiate the identity $g^{''} = 2 (g^{'} - g)$ twice, to get $g^{'''} = 2 (g^{''} - g^{'})$ and then

g^{'} (x) = e^{x} (cos x - sin x), g^{''} (x) = e^{x} (cos x - sin x) + e^{x} (- sin x - cos x) = - 2 e^{x} sin x 2 [e^{x} (cos x - sin x) - e^{x} cos x] = - 2 e^{x} sin x = g^{''} (x) g^{(4)} (x) = 2 (g^{'''} (x) - g^{''} (x))

(d)

5 marks

At $x = 0$ : $g (0) = 1$ , $g^{'} (0) = 1$ , and $g^{''} (0) = 2 (1 - 1) = 0$ . Then $g^{'''} (0) = 2 (0 - 1) = - 2$ and $g^{(4)} (0) = 2 (- 2 - 0) = - 4$ .

g (x) \approx 1 + x + \frac{0}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 4}{4 !} x^{4} = 1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4}

(e)

3 marks

Use part (d): $e^{x} cos x - 1 - x = - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots$ .

x \to 0 lim \frac{e ^{x} cos x - 1 - x}{x ^{3}} = x \to 0 lim (- \frac{1}{3} - \frac{1}{6} x + \dots) = - \frac{1}{3}

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra45 marks · 41%
Q1, Q4, Q6, Q8, Q9, Q10
Calculus29 marks · 26%
Q7, Q12
Geometry and trigonometry15 marks · 14%
Q11
Functions14 marks · 13%
Q3, Q5
Statistics and probability7 marks · 6%
Q2

Mentor insights

Traps to watch for

01
The two lines cross at the means · Q2
The regression line of $a$ on $c$ and the regression line of $c$ on $a$ meet at $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Solving them together gives the mean ages.
02
Total area is 1 · Q7
The density must integrate to 1 over $0 \leq x \leq 1$ . The antiderivative is an inverse sine, so be ready with $arcsin \frac{3}{2} = \frac{π}{3}$ .
03
Use the definition of each type · Q10
A geometric series has a constant ratio of consecutive terms, an arithmetic series a constant difference. Write the equation for equal ratios (or differences) and cancel $ln x$ .
04
No solution because of parallelism · Q11
Two planes meet in a line. If that line is parallel to the third plane but not inside it, the three planes have no common point.
05
Use the result from the previous part · Q12
Part (c) hands you a relationship between derivatives. In (d), use it to find the derivatives at 0 without differentiating a fourth time.

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2022 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 56 marks

Question 1

5 marks

A definite integral with a square root

Number and algebra

5 marks

Split the integrand.
Integrate term by term.
Evaluate between 1 and 9.

\frac{3 x - 5}{x} = 3 - 5 x^{- 1/2} \int (3 - 5 x^{- 1/2}) d x = 3 x - 10 x^{1/2} + c [3 x - 10 x]_{1}^{9} = (27 - 30) - (3 - 10) = - 3 + 7 = 4

Question 2

7 marks

A box plot, outliers and two regression lines

Statistics and probability

(a)

3 marks

The quartiles are $Q_{1} = 6$ and $Q_{3} = 10$ , so the IQR is 4. A value is an outlier above $Q_{3} + 1.5 \times IQR$ .
So the largest value that is not an outlier is $c = 16$ .

10 + 1.5 \times 4 = 16

(b)

4 marks

(i) To estimate the child's age from the adult's, use the regression line of $c$ on $a$ .

(ii) Both lines pass through $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Substitute $c = \frac{1}{2} a - 9$ into $a = \frac{7}{4} c + 20$ .

The mean age of the adults is 34 years.

c = \frac{1}{2} (42) - 9 = 12 years a = \frac{7}{4} (\frac{1}{2} a - 9) + 20 = \frac{7}{8} a + 4.25 \Rightarrow \frac{1}{8} a = 4.25 \Rightarrow a = 34

Question 3

7 marks

A composite of trig functions and an equation

Functions

(a)

2 marks

Replace $x$ by $2 x$ in $f$ .

(f \circ g) (x) = f (2 x) = 3 sin 2 x + cos 2 x

(b)

5 marks

Set $3 sin 2 x + cos 2 x = 2 cos 2 x$ , so $3 sin 2 x = cos 2 x$ .
For $0 \leq x \leq π$ we need $2 x$ in $[0, 2 π]$ : the tangent equals $\frac{1}{3}$ at $\frac{π}{6}$ and $\frac{7 π}{6}$ .

tan 2 x = \frac{1}{3} 2 x = \frac{π}{6}, \frac{7 π}{6} \Rightarrow x = \frac{π}{12}, \frac{7 π}{12}

Question 4

5 marks

A tangent parallel to a given line

Number and algebra

5 marks

Use the product rule on $y = (2 x - 1) e^{k x}$ .
At $x = 1$ the gradient is $2 e^{k} + k e^{k} = (k + 2) e^{k}$ . The line is $y = 5 e^{k} x$ , with gradient $5 e^{k}$ . Set them equal and divide by $e^{k}$ .

\frac{d y}{d x} = 2 e^{k x} + k (2 x - 1) e^{k x} (k + 2) e^{k} = 5 e^{k} \Rightarrow k + 2 = 5 \Rightarrow k = 3

Question 5

7 marks

Describing two transformations, then the smallest $y$ -intercept

Functions

(a)

2 marks

(b)

5 marks

Every value of $g$ is at least $7$ . The least value of the sine term is $- 4$ , so
The $y$ -intercept is $g (0)$ . Since $sin (- \frac{3 π}{2}) = 1$ :
The smallest value of $r$ comes from the smallest $q$ .

- 4 + 2.5 + q \geq 7 \Rightarrow q \geq 8.5 r = 4 (1) + 2.5 + q = 6.5 + q r_{m i n} = 6.5 + 8.5 = 15

Question 6

5 marks

The constant term of a binomial expansion

Number and algebra

5 marks

The general term of $(8 x^{3} - \frac{1}{2 x})^{n}$ is
The power of $x$ is $3 (n - r) - r = 3 n - 4 r$ . For a constant term this must be 0.
$r$ must be a whole number from 0 to $n$ , so $n$ must be a multiple of 4. The coefficient is then non-zero.

T_{r + 1} = (r n) (8 x^{3})^{n - r} (- \frac{1}{2 x})^{r} 3 n - 4 r = 0 \Rightarrow r = \frac{3 n}{4} n = 4, 8, 12, \dots (n a positive multiple of 4)

Question 7

8 marks

A density function with a square root: finding $k$ and the mean

Calculus

(a)

4 marks

The integral of $\frac{1}{4 - 3 x ^{2}}$ is $\frac{1}{3} arcsin \frac{3 x}{2}$ . Set the total area to 1.

\int_{0}^{1} \frac{k}{4 - 3 x ^{2}} d x = \frac{k}{3} arcsin \frac{3}{2} = \frac{k}{3} \cdot \frac{π}{3} = 1 k = \frac{3 3}{π}

(b)

4 marks

$E (X) = \int_{0}^{1} \frac{k x}{4 - 3 x ^{2}} d x$ . The numerator is nearly the derivative of the inside: $\frac{d}{d x} (4 - 3 x^{2}) = - 6 x$ .

E (X) = k [- \frac{1}{3} 4 - 3 x^{2}]_{0}^{1} = \frac{k}{3} (2 - 1) = \frac{k}{3} = \frac{3}{π}

Question 8

6 marks

Proof by contradiction: two odd numbers can't have squares summing to a multiple of 4

Number and algebra

6 marks

Assume that $a$ and $b$ are both odd, so $a = 2 m + 1$ and $b = 2 n + 1$ for some integers $m$ and $n$ .
This is a multiple of 4 plus 2, so it leaves remainder 2 when divided by 4. It is not divisible by 4. That contradicts the assumption that $a^{2} + b^{2}$ is exactly divisible by 4.
So $a$ and $b$ cannot both be odd.

a^{2} + b^{2} = (4 m^{2} + 4 m + 1) + (4 n^{2} + 4 n + 1) = 4 (m^{2} + m + n^{2} + n) + 2

Question 9

6 marks

Two complex numbers: their product and an argument condition

Number and algebra

(a)

3 marks

Multiply out and collect the real and imaginary parts.

z_{1} z_{2} = (1 + b i) ((1 - b^{2}) - 2 b i) = (1 - b^{2}) - 2 b i + b (1 - b^{2}) i - 2 b^{2} i^{2} z_{1} z_{2} = (1 + b^{2}) + i (- b - b^{3}) = (1 + b^{2}) (1 - b i)

(b)

3 marks

The argument is $arctan (\frac{- b ( 1 + b ^{2} )}{1 + b ^{2}}) = arctan (- b)$ , since $1 + b^{2} \neq = 0$ .

arctan (- b) = \frac{π}{4} \Rightarrow - b = 1 \Rightarrow b = - 1

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

18 marks

A series that is geometric, and then arithmetic

Number and algebra

(a)

6 marks

The terms are $ln x$ , $p ln x$ , $\frac{1}{3} ln x$ . (i) For a geometric series the ratios are equal.
(ii) The common ratio is $r = p = \pm \frac{1}{3}$ , so $∣ r ∣ < 1$ and the series converges.
(iii) For $p > 0$ , $r = \frac{1}{3}$ , and the sum to infinity is

\frac{p ln x}{ln x} = \frac{\frac{1}{3} ln x}{p ln x} \Rightarrow p = \frac{1}{3 p} \Rightarrow p^{2} = \frac{1}{3} \Rightarrow p = \pm \frac{1}{3} S_{\infty} = \frac{ln x}{1 - \frac{1}{3}} = 3 + 3 \Rightarrow ln x = (3 + 3) (1 - \frac{1}{3}) = 2 x = e^{2}

(b)

12 marks

(i) For an arithmetic series consecutive differences are equal.

(ii) The common difference is the first term minus the second, with the sign reversed.

(iii) The sum of the first $n$ terms is $S_{n} = \frac{n}{2} [2 ln x + (n - 1) d]$ , and it equals $ln (\frac{1}{x ^{3}}) = - 3 ln x$ .

p ln x - ln x = \frac{1}{3} ln x - p ln x \Rightarrow 2 p = \frac{4}{3} \Rightarrow p = \frac{2}{3} d = p ln x - ln x = - \frac{1}{3} ln x \frac{n}{2} [2 - \frac{n - 1}{3}] ln x = - 3 ln x \Rightarrow \frac{n ( 7 - n )}{6} = - 3 n^{2} - 7 n - 18 = 0 \Rightarrow (n - 9) (n + 2) = 0 \Rightarrow n = 9

Question 11

15 marks

Three planes with no common point, a line and a distance

Geometry and trigonometry

(a)

4 marks

The direction of the line where $Π_{1}$ and $Π_{2}$ meet is the cross product of their normals.
This direction is perpendicular to the normal of $Π_{3}$ , since $(- 9) (- 1) + 3 (- 5) + (- 2) (- 3) = 9 - 15 + 6 = 0$ . So the line is parallel to $Π_{3}$ .
A point on the line is $(1, - 2, 0)$ . It is not in $Π_{3}$ : $- 9 - 6 - 0 = - 15 \neq = 32$ . So the line never meets $Π_{3}$ , and the three planes have no common point.

2 - 1 1 \times 1 - 2 3 = - 1 - 5 - 3

(b)

5 marks

(i) For $P (1, - 2, 0)$ :

(ii) Use P and the direction vector from part (a).

Π_{1} : 2 + 2 + 0 = 4, Π_{2} : 1 + 4 + 0 = 5 r = 1 - 2 0 + λ 153

(c)

6 marks

The line is parallel to the plane, so every point of L is the same distance from $Π_{3}$ . Use the distance from P to $Π_{3}$ , whose normal is $(- 9, 3, - 2)$ with $∣ n ∣ = 94$ .

d = \frac{∣ - 9 ( 1 ) + 3 ( - 2 ) - 2 ( 0 ) - 32∣}{81 + 9 + 4} = \frac{47}{94} = \frac{94}{2} \approx 4.85

Question 12

21 marks

Maclaurin series of $e^{x} sin x$ and $e^{x} cos x$ , and a limit

Calculus

(a)

4 marks

Multiply the series for $e^{x}$ and $sin x$ and keep the terms up to $x^{3}$ .

(1 + x + \frac{x ^{2}}{2}) (x - \frac{x ^{3}}{6}) = x + x^{2} + \frac{x ^{3}}{2} - \frac{x ^{3}}{6} = x + x^{2} + \frac{1}{3} x^{3}

(b)

4 marks

Replace $x$ by $x^{2}$ in the series for $f$ , then integrate term by term.

\int_{0}^{1} e^{x^{2}} sin (x^{2}) d x \approx \int_{0}^{1} (x^{2} + x^{4} + \frac{1}{3} x^{6}) d x = \frac{1}{3} + \frac{1}{5} + \frac{1}{21} = \frac{61}{105}

(c)

5 marks

(i) Differentiate $g (x) = e^{x} cos x$ with the product rule twice.

Compare with $2 (g^{'} (x) - g (x))$ .

(ii) Differentiate the identity $g^{''} = 2 (g^{'} - g)$ twice, to get $g^{'''} = 2 (g^{''} - g^{'})$ and then

g^{'} (x) = e^{x} (cos x - sin x), g^{''} (x) = e^{x} (cos x - sin x) + e^{x} (- sin x - cos x) = - 2 e^{x} sin x 2 [e^{x} (cos x - sin x) - e^{x} cos x] = - 2 e^{x} sin x = g^{''} (x) g^{(4)} (x) = 2 (g^{'''} (x) - g^{''} (x))

(d)

5 marks

At $x = 0$ : $g (0) = 1$ , $g^{'} (0) = 1$ , and $g^{''} (0) = 2 (1 - 1) = 0$ . Then $g^{'''} (0) = 2 (0 - 1) = - 2$ and $g^{(4)} (0) = 2 (- 2 - 0) = - 4$ .

g (x) \approx 1 + x + \frac{0}{2 !} x^{2} + \frac{- 2}{3 !} x^{3} + \frac{- 4}{4 !} x^{4} = 1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4}

(e)

3 marks

Use part (d): $e^{x} cos x - 1 - x = - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots$ .

x \to 0 lim \frac{e ^{x} cos x - 1 - x}{x ^{3}} = x \to 0 lim (- \frac{1}{3} - \frac{1}{6} x + \dots) = - \frac{1}{3}

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra45 marks · 41%
Q1, Q4, Q6, Q8, Q9, Q10
Calculus29 marks · 26%
Q7, Q12
Geometry and trigonometry15 marks · 14%
Q11
Functions14 marks · 13%
Q3, Q5
Statistics and probability7 marks · 6%
Q2

Mentor insights

Traps to watch for

01
The two lines cross at the means · Q2
The regression line of $a$ on $c$ and the regression line of $c$ on $a$ meet at $(\overset{c}{ˉ}, \overset{a}{ˉ})$ . Solving them together gives the mean ages.
02
Total area is 1 · Q7
The density must integrate to 1 over $0 \leq x \leq 1$ . The antiderivative is an inverse sine, so be ready with $arcsin \frac{3}{2} = \frac{π}{3}$ .
03
Use the definition of each type · Q10
A geometric series has a constant ratio of consecutive terms, an arithmetic series a constant difference. Write the equation for equal ratios (or differences) and cancel $ln x$ .
04
No solution because of parallelism · Q11
Two planes meet in a line. If that line is parallel to the third plane but not inside it, the three planes have no common point.
05
Use the result from the previous part · Q12
Part (c) hands you a relationship between derivatives. In (d), use it to find the derivatives at 0 without differentiating a fourth time.

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