## Download our Successful College Application Guide

Our Guide is written by counselors from **Cambridge University** for colleges like **MIT** and other** Ivy League colleges.**

To join our college counseling program, call at +918825012255

Skip to content
## IB Math AA HL Paper 1 Question Bank

**1.) A fair die is thrown 3 times. Let X be the number of throws resulting in a six.**

**2.) Given that = 2 â€“ i, z, find z in the form a + ib.**

**3.) Consider the cubic function f(x) = ax**^{3} + 2x^{2} + 3x + 4. Find the value of a in each of the following cases.

**4.) Solve the equations**

**5.) Solve the equation tan**^{2}x = 3 for -Ï€â‰¤ x â‰¤ Ï€

**6.) Solve the following system of equations**

**7.) A fair six-sided die, with sides numbered 1, 1, 2,3, 4, 5 is thrown. Find the mean and variance of the score.**

## Download our Successful College Application Guide

**8.) Let f(z) = z**^{2} – 8z + 20

**9.) Given the function f(x) = x**^{2} – 3bx + (c+2), determine the values of b and c such as f(1) = 0 and fâ€™(3) = 0.

**10.) If ln(2x-1), find d**^{2}y/dx^{2}

**11.) Find âˆ«((3x**^{2} – x + 2âˆšx)/3âˆšx)dx

**12.) Consider the function f(x) = 2×3 + 3×2 – 12x + 5.**

## Our Expert Tutors!

### Amrita

### Aiman

### Jyoti

### Vaishnavi Thakur

### SOURAV

Edit Template
## Get access to our free IB resources

### IBDP Study Notes

### IB Comprehensive Syllabus

### IB IA Ideas

### IB CAS Ideas

### IB Extended Essay Ideas

Edit Template
#### We Are Here To Help You To Excel in Your Exams!

### Book Your Free Demo Session Now!

#### Head Office

#### International IB Tutors

###### â’¸ 2023 TYCHR ACADEMY | All Rights Reserved

If you’re looking for a comprehensive IB Math AA HL Paper 1 Question Bank, look no further! This resource contains hundreds of practice questions, organized by topic, to help you review for your upcoming test. Whether you’re struggling with algebra or just want to brush up on your trigonometry, this Question Bank has you covered. With answer explanations and step-by-step worked solutions included for every question, you can be confident that you’re getting the best possible preparation for your IB Math AA HL Paper 1 exam.

**Section A**

**a) Write down the probability mass function of X.**

The probability mass function of X is given by the binomial probability distribution. In this case, the number of trials is 3, and the probability of success in each trial is 1/6, since it is a fair die. So the probability mass function of X is:

P(X = x) = (3 choose x) * (1/6)^{x} * (5/6)^{(3-x)} for x = 0, 1, 2, 3

**b) Find the expected value of X.**

To find the expected value of X, we use the formula E(X) = âˆ‘xP(X = x) for x = 0, 1, 2, 3

E(X) = (0 * (3 choose 0) * (1/6)^{0}Â * (5/6)^{3}) + (1 * (3 choose 1) * (1/6)^{1}Â * (5/6)^{2}) + (2 * (3 choose 2) * (1/6)^{2} * (5/6)^{1}) + (3 * (3 choose 3) * (1/6)^{3} * (5/6)^{0})

E(X) = 3/2

z = (2 â€“ i)(z + 2)

= 2z + 4 â€“ iz â€“ 2i

z(1 â€“ i) = â€“ 4 + 2i

z = â€“ 3 â€“ i

**a) the graph of the function passes through the point (1,10).**

f(1)= 10, a=1

**b) f(x) is divisible by (x-1)**

f(1) = 0, a= -9

**c) when f(x) is divided by (x-1), the remainder in the 10**

f(1) = 10, a=1

**a) logx + log(x+1) = log6**

x(x+1)=6

x^{2}+x-6=0

x=2

**b) logx + log(x+3) = 10**

x(x+3)=10

x^{2}+3x-10=0

x=2

**c) log(x+18) – logx = 1**

(x+18)/x= 1

tanx = Â±âˆš3

For tanx = âˆš3

x=Ï€/3, x= -2Ï€/3

For tanx = -âˆš3

x=-Ï€/3, x= 2Ï€/3

X + 3y – 2z = -6

2x + y + 3z = 7

3x – y + z = 6

Back substitution gives x = 1, y = -1, z = 2.

Mean = â…™ (1+1+2+3+4+5) = 8/3

Variance= â…™ (1+1+4+9+16+25) – 64/9 = 20/9

Our Guide is written by counselors from **Cambridge University** for colleges like **MIT** and other** Ivy League colleges.**

To join our college counseling program, call at +918825012255

**a) Find the discriminant of the quadratic function f**

Discriminant= -16

**b) Find the complex roots of equation f(z) = 0 in the form z = x Â± yi**

z= (8Â±4i)/ 2 = 4Â±2i

**c) Use factorisation to express f in the form of f(z) = (z-h) ^{2}+k**

(z-4-2i)(z-4-2i)

= (z-4)

f(1) = 1 – 3b + c + 2 = 0

fâ€™(x) = 2x – 3b

fâ€™(3) = 6 – 3b = 0

b = 2

1 – 3(2) + c + 2 = 0

c =3

dy/dx = 2/2x-1

d^{2}y/dx^{2 }= 2(2x-1)^{-2}(2)

d^{2}y/dx^{2 }= -4/(2x-1)^{2}

= âˆ«(x^{1.5 }– â…“ x^{0.5 }+ â…” )dx = (x^{2.5}/2.5) – (x^{1.5}/4.5) + (â…”)x + c

**Section B**

**a) Find the stationary points of f(x) and determine their nature (whether they are local maximum, local minimum, or neither).**

Solution:

Finding stationary points and their nature:

Step 1: Find the first derivative of f(x)

f'(x) = 6x^{2} + 6x – 12

Step 2: Set f'(x) = 0 to find stationary points

6x^{2} + 6x – 12 = 0

Step 3: Solve for x

Dividing both sides by 6, we get:

x^{2}+ x – 2 = 0

Factoring the quadratic equation, we get:

(x + 2)(x – 1) = 0

Setting each factor to zero, we get two possible values for x:

x + 2 = 0 or x – 1 = 0

Solving for x, we get:

x = -2 or x = 1

So, the stationary points of f(x) are at x = -2 and x = 1.

Step 4: Determine the nature of the stationary points

To determine the nature of the stationary points, we can use the second derivative test. Taking the second derivative of f(x):

f”(x) = 12x + 6

Evaluating f”(x) at x = -2 and x = 1:

f”(-2) = 12(-2) + 6 = -18

f”(1) = 12(1) + 6 = 18

Since f”(-2) is negative, the stationary point at x = -2 is a local maximum. Since f”(1) is positive, the stationary point at x = 1 is a local minimum.

**b) Find the equation of the tangent line to the curve of f(x) at the point where x = 2.**

Solution:

Finding the equation of the tangent line at x = 2:

Step 1: Find the slope of the tangent line

The slope of the tangent line to the curve of f(x) at any point x can be found using the first derivative f'(x). So, we can evaluate f'(2):

f'(2) = 6(2)^{2} + 6(2) – 12 = 48

So, the slope of the tangent line at x = 2 is 48.

Step 2: Find the point of tangency

The point of tangency is the point where x = 2, so we can plug x = 2 into the original function f(x):

f(2) = 2(2)^{3} + 3(2)2^{2} – 12(2) + 5 = 37

So, the point of tangency is (2, 37).

Step 3: Write the equation of the tangent line

Using the slope and the point of tangency, we can write the equation of the tangent line in point-slope form:

y – y_{1} = m(x – x_{1})

Plugging in the values:

y – 37 = 48(x – 2)

Simplifying and rearranging, we get:

y = 48x – 95

So, the equation of the tangent line to the curve of f(x) at the point where x = 2 is y = 48x – 95.

**c) Finding the area under the curve of f(x) between x = 0 and x = 4:**

Solution:

Step 1: Find the indefinite integral of f(x)

The indefinite integral of f(x) is given by:

âˆ«f(x)dx = âˆ«(2x^{3}+ 3x^{2} – 12x + 5)dx

Integrating each term separately:

âˆ«2x^{3}dx + âˆ«3x^{2} dx – âˆ«12xdx + âˆ«5dx

= (1/2)x^{4} + x^{3} – 6x^{2} + 5x + C

where C is the constant of integration.

Step 2: Evaluate the definite integral

To find the area under the curve of f(x) between x = 0 and x = 4, we can evaluate the definite integral:

âˆ«[f(x)]dx from x = 0 to x = 4

= [(1/2)(4)^{4} + (4)^{3} – 6(4)^{2} + 5(4)] – [(1/2)(0)^{4} + (0)^{3} – 6(0)^{2} + 5(0)]

= 128 + 64 – 96 + 20

= 116

So, the area under the curve of f(x) between x = 0 and x = 4 is 116 square units.

**d) Finding the x-coordinate of the point where the curve of f(x) has a horizontal tangent:**

Solution:

Step 1: Find the critical points

The points where the curve of f(x) has a horizontal tangent are the points where the first derivative f'(x) = 0. From part (a), we already found the first derivative f'(x) = 6x^{2} + 6x – 12. Setting f'(x) = 0 and solving for x:

6x^{2} + 6x – 12 = 0

Dividing both sides by 6, we get:

x^{2} + x – 2 = 0

Factoring the quadratic equation, we get:

(x + 2)(x – 1) = 0

Setting each factor to zero, we get two possible values for x:

x + 2 = 0 or x – 1 = 0

Solving for x, we get:

x = -2 or x = 1

So, the critical points of f(x) are at x = -2 and x = 1.

Step 2: Determine the x-coordinate of the point with a horizontal tangent

Since the x-coordinate of the point with a horizontal tangent is the point where the first derivative f'(x) = 0, we can see that x = -2 is the x-coordinate of the point where the curve of f(x) has a horizontal tangent.

So, the x-coordinate of the point where the curve of f(x) has a horizontal tangent is x = -2.

**e) Finding the x-values where the curve of f(x) is concave up or concave down:**

Solution:

Step 1: Find the second derivative of f(x)

The second derivative of f(x) is given by:

f”(x) = (f'(x))’

From part (a), we already found the first derivative f'(x) = 6x^{2} + 6x – 12. Taking the derivative of f'(x), we get:

f”(x) = (6x^{2} + 6x – 12)’

Using the power rule and sum rule of differentiation, we get:

f”(x) = 12x + 6

Step 2: Determine the concavity of the curve

The concavity of the curve of f(x) is determined by the sign of the second derivative f”(x). If f”(x) > 0, then the curve is concave up. If f”(x) < 0, then the curve is concave down.

Setting f”(x) = 0 and solving for x:

12x + 6 = 0

Dividing both sides by 12, we get:

x + 1/2 = 0

Solving for x, we get:

x = -1/2

So, the curve of f(x) changes concavity at x = -1/2. For x < -1/2, the curve is concave down, and for x > -1/2, the curve is concave up.

**f) Finding the inflection points of the curve:**

Solution:

Step 1: Determine the x-coordinate of the inflection point

Since the inflection point is the point where the curve changes concavity, we can see from part (e) that the curve changes concavity at x = -1/2. Therefore, the x-coordinate of the inflection point is x = -1/2.

Step 2: Find the y-coordinate of the inflection point

To find the y-coordinate of the inflection point, we can substitute x = -1/2 into the original function f(x):

f(-1/2) = 2(-1/2)^{3} + 3(-1/2)^{3} – 6(-1/2) + 5

Simplifying, we get:

f(-1/2) = -1/8 + 3/4 + 3 + 5

f(-1/2) = 25/8

So, the y-coordinate of the inflection point is y = 25/8.

Therefore, the inflection point of the curve of f(x) is at (-1/2, 25/8).

IBDP Maths AIhl

ap statistics

IBDP Biology

ap Biology

IBDP Business Management

ap microeconomics

IBDP ESS

Ap Environmental Science

IBDP Physics

ap physics c mechanics

- HD-213, WeWork DLF Forum, Cyber City, DLF Phase 3, DLF, Gurugram, Haryana-122002
- +919540653900
- contact@tychr.com

Free Trial Class!

- Login
- Sign Up

SUCCESSFUL COLLEGE APPLICATIONS GUIDE RELEASED

**Download our Successful College Application Guide** developed by counselors from the **University of Cambridge** for institutions like **Oxbridge**Â alongside other **Ivy Leagues**. To join our college counseling program, call at **+918825012255**

WE ARE HIRING

We are hiring a Business Development Associate and Content Writer and Social Media Strategist at our organisation TYCHR to take over the responsibility of conducting workshops and excelling in new sales territory. **View More**

For all the results, Please Click here

Join the League

Please fill out the form below for a free demo class!

0

0

Your Cart
Your cart is empty