Free IB Maths AA SL and HL worked questions

SL and HL · no calculator

6 marksPaper 1 · no calculatorSL and HL

The curve

C

has equation

y = x^{2} - 3x + 1

. The point

A(1, -1)

lies on

C

.

(a)

Find the equation of the normal to

C

at

A

.

[4]

(b)

The normal meets

C

again at

B

. Find the coordinates of

B

.

[2]

See the complete 6-mark scheme

Part (a) · 4 marks

Differentiate and substitute $x = 1$ .
$\frac{dy}{dx} = 2x - 3 = -1 \text{ at } x = 1$
M1Differentiating and substituting $x = 1$ .Lost if: Substituting into the curve's equation instead of the derivative.
A1Tangent gradient $-1$ .Lost if: Arithmetic slip, $2 - 3 = 1$ .Needs M1 (a) first.
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $-1$ .
$m_{\text{normal}} = 1$
(M1)Using the negative reciprocal of their tangent gradient.Lost if: Using the tangent gradient itself for the normal.Implied: awarded if later correct work shows it.
Use the point $A(1, -1)$ .
$y + 1 = 1(x - 1) \Rightarrow y = x - 2$
A1 $y = x - 2$ .Lost if: Writing the tangent $y = -x$ instead of the normal.Needs (M1) (a) first.

Part (b) · 2 marks

Solve the curve and the normal simultaneously.
$x^{2} - 3x + 1 = x - 2 \Rightarrow x^{2} - 4x + 3 = 0$
M1Equating the curve and their normal.Lost if: Equating the curve to the tangent.
Factorise; $x = 1$ is $A$ .
$(x - 1)(x - 3) = 0 \Rightarrow x = 3,\ y = 1$
A1 FT $(3, 1)$ .Lost if: Giving $(1, -1)$ , which is $A$ itself.Needs M1 (b) first. Follow-through from part (a): Follow their normal if it is a straight line through $A$ .

HL Paper 3 · calculator

24 marksPaper 3 · HL problem solvingHL only

For integers

n\ge0

, let

I_n=\displaystyle\int_0^1x^ne^x\,dx

. The graph shows

y=e^x

on

0\le x\le1

.

(a)

Find the exact values of

I_{0}

and

I_{1}

.

[3]

(b)

Show that

I_{n} = e - nI_{n-1}

for

n \ge 1

.

[3]

(c)

Hence find

I_{2}

and

I_{3}

in the form

p + qe

, where

p, q \in \mathbb{Z}

.

[2]

(d)

Use your GDC to find the value of

nI_{n}

for

n = 10

,

n = 100

and

n = 1000

. Hence conjecture the value of

\displaystyle\lim_{n \to \infty} nI_{n}

.

[2]

(e)

Prove that

\dfrac{1}{n + 1} \le I_{n} \le \dfrac{e}{n + 1}

for all

n \in \mathbb{N}

.

[4]

(f)

Hence write down

\displaystyle\lim_{n \to \infty} I_{n}

.

[1]

(g)

Use part (b) and part (f) to prove the conjecture in part (d).

[3]

(hi)

Extend the investigation: let

J_{n} = \displaystyle\int_{0}^{1} x^{n}e^{-x}\,dx

. Show that

J_{n} = nJ_{n-1} - e^{-1}

for

n \ge 1

.

[2]

(hii)

Prove by mathematical induction that

J_{n} = n!\left(1 - e^{-1}\displaystyle\sum_{k=0}^{n} \frac{1}{k!}\right)

for all

n \in \mathbb{N}

.

[4]

See the complete 24-mark scheme

Part (a) · 3 marks

$I_{0} = \int_{0}^{1} e^{x}\,dx = e - 1$ .
A1 $I_{0} = e - 1$ .Lost if: Writing $e$ , forgetting the lower limit gives $e^{0} = 1$ .
By parts with $u = x$ , $\frac{dv}{dx} = e^{x}$ .
$I_{1} = \left[xe^{x}\right]_{0}^{1} - \int_{0}^{1} e^{x}\,dx = e - (e - 1) = 1$
M1Integration by parts on $I_{1}$ with $u = x$ .Lost if: Choosing $u = e^{x}$ , which raises the power of $x$ .
A1 $I_{1} = 1$ .Lost if: Sign slip on the lower-limit bracket.Needs M1 (a) first.

Part (b) · 3 marks

Take $u = x^{n}$ and $\frac{dv}{dx} = e^{x}$ , so $\frac{du}{dx} = nx^{n-1}$ and $v = e^{x}$ .
$I_{n} = \left[x^{n}e^{x}\right]_{0}^{1} - \int_{0}^{1} nx^{n-1}e^{x}\,dx$
M1Parts with $u = x^{n}$ and $v = e^{x}$ .Lost if: Working backwards from the given recurrence with $n = 2$ : a check of one case is not a proof for all $n$ .
A1Correct bracket and integral, $\left[x^{n}e^{x}\right]_{0}^{1} - n\int_{0}^{1} x^{n-1}e^{x}\,dx$ .Lost if: Losing the factor $n$ .Needs M1 (b) first.
The bracket is $e - 0$ for $n \ge 1$ , and the integral is $nI_{n-1}$ .
$I_{n} = e - nI_{n-1}$
A1Bracket evaluated as $e$ (the lower limit gives 0 because $n \ge 1$ ) and the integral named $nI_{n-1}$ .Lost if: Evaluating the bracket as $e - 1$ .Needs M1 (b) first.
AG $I_{n} = e - nI_{n-1}$ .The answer was given in the question, so writing it earns nothing: the marks are for the working that reaches it.

Part (c) · 2 marks

Use the recurrence twice.
$I_{2} = e - 2I_{1} = e - 2, \quad I_{3} = e - 3(e - 2) = 6 - 2e$
A1 $I_{2} = e - 2$ .Lost if: Using $I_{2} = e - 2I_{0}$ .
A1 FT $I_{3} = 6 - 2e$ .Lost if: Sign slip expanding $-3(e - 2)$ .Follow-through from part (a): Follow their $I_{1}$ .

Part (d) · 2 marks

Use GDC definite integration for each value. The forward recurrence amplifies rounding errors and is unsuitable for these large indices.
$10I_{10} = 2.2800\ldots,\quad 100I_{100} = 2.6652\ldots,\quad 1000I_{1000} = 2.7129\ldots$
A1At least two of the three values correct to 3 sf.Lost if: Computing $I_{n}$ without multiplying by $n$ .
The values increase towards $2.718\ldots$ , which suggests the limit is $e$ .
A1Conjecture: the limit is $e$ .Lost if: Giving $2.71$ or $2.7$ as the limit: name the constant the numbers suggest.

Part (e) · 4 marks

For $0 \le x \le 1$ , $1 \le e^{x} \le e$ , and $x^{n} \ge 0$ , so multiplying keeps the order.
$x^{n} \le x^{n}e^{x} \le ex^{n}$
M1Bounding $e^{x}$ between 1 and $e$ on $[0, 1]$ and multiplying by $x^{n}$ .Lost if: Bounding $x^{n}e^{x}$ by its values at the endpoints only.
R1Multiplying by $x^{n} \ge 0$ keeps the inequality directions.Lost if: Not stating why the direction is kept.
Integrating over $[0, 1]$ keeps the inequalities.
$\int_{0}^{1} x^{n}\,dx \le I_{n} \le e\int_{0}^{1} x^{n}\,dx$
R1Integrating an inequality over $[0, 1]$ preserves it.Lost if: Integrating only one side.
Evaluate.
$\frac{1}{n + 1} \le I_{n} \le \frac{e}{n + 1}$
A1Both integrals evaluated, $\frac{1}{n + 1}$ and $\frac{e}{n + 1}$ .Lost if: Writing $\frac{1}{n}$ .Needs M1 (e) first.

Part (f) · 1 marks

Both bounds tend to 0, so $I_{n}$ is squeezed to 0.
A1 $0$ with reference to both bounds tending to 0.Lost if: Using only the lower bound.

Part (g) · 3 marks

Rearrange the recurrence.
$nI_{n-1} = e - I_{n} \to e - 0 = e$
M1Rearranging the recurrence to $nI_{n-1} = e - I_{n}$ .Lost if: Using the numerical values from part (d) as the proof: evidence is not proof.
R1 $I_{n} \to 0$ used, so $nI_{n-1} \to e$ .Lost if: Not quoting part (f).
Replace $n$ by $n + 1$ : $(n + 1)I_{n} \to e$ .
$nI_{n} = \frac{n}{n + 1}(n + 1)I_{n} \to 1 \times e = e$
R1Shifting the index and handling $\frac{n}{n + 1} \to 1$ .Lost if: Treating $nI_{n-1}$ and $nI_{n}$ as the same sequence without comment.

Part (hi) · 2 marks

Parts with $u = x^{n}$ , $\frac{dv}{dx} = e^{-x}$ , $v = -e^{-x}$ .
$J_{n} = \left[-x^{n}e^{-x}\right]_{0}^{1} + n\int_{0}^{1} x^{n-1}e^{-x}\,dx = -e^{-1} + nJ_{n-1}$
M1Parts with $v = -e^{-x}$ .Lost if: Taking $v = e^{-x}$ (sign).
A1Correct bracket $-e^{-1}$ and $+nJ_{n-1}$ .Lost if: Sign slip on the integral term.Needs M1 (hi) first.
AG $J_{n} = nJ_{n-1} - e^{-1}$ .The answer was given in the question, so writing it earns nothing: the marks are for the working that reaches it.

Part (hii) · 4 marks

Base case $n = 0$ : $J_{0} = \int_{0}^{1} e^{-x}\,dx = 1 - e^{-1}$ , and the formula gives $0!(1 - e^{-1} \cdot 1)$ . True.
A1Base case checked at $n = 0$ (or $n = 1$ ), both sides evaluated.Lost if: Stating "true for $n = 0$ " without evaluating either side.
Assume true for $n = m$ . Then, by part (h)(i),
$J_{m+1} = (m + 1)J_{m} - e^{-1} = (m + 1)!\left(1 - e^{-1}\sum_{k=0}^{m} \frac{1}{k!}\right) - e^{-1}$
M1Assumption for $n = m$ stated, and the recurrence used for $J_{m+1}$ .Lost if: Assuming the result for $n = m + 1$ .
Take out $(m + 1)!$ : the last term is $(m + 1)! \times e^{-1}\frac{1}{(m + 1)!}$ .
$J_{m+1} = (m + 1)!\left(1 - e^{-1}\sum_{k=0}^{m+1} \frac{1}{k!}\right)$
A1Writing $e^{-1}$ as $(m + 1)!\,e^{-1}\frac{1}{(m + 1)!}$ to join the sum.Lost if: Leaving $-e^{-1}$ outside the bracket.Needs M1 (hii) first.
True for $n = 0$ , and true for $n = m$ implies true for $n = m + 1$ , so true for all $n \in \mathbb{N}$ by induction.
R1Complete concluding statement linking base case and inductive step.Lost if: No conclusion, or a conclusion that does not mention both steps.

A conjecture still needs proof

“The values approach e, so the limit is e” earns no proof marks. A finite table cannot establish a limit.

The recurrence gives (n + 1)Iₙ = e − Iₙ₊₁. Since Iₙ₊₁ tends to 0, (n + 1)Iₙ tends to e. Multiplying by n/(n + 1), which tends to 1, proves nIₙ tends to e.

Try another investigation Understand Paper 3 and practise a first move Practise using the AA booklet

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