IB Math AA HL Paper 1 Question Bank

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

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

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) = ∑xP(X = x) for x = 0, 1, 2, 3
E(X) = (0 * (3 choose 0) * (1/6)0 * (5/6)3) + (1 * (3 choose 1) * (1/6)1 * (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

2.) Given that = 2 – i, z, find z in the form a + ib.

z = (2 – i)(z + 2)
= 2z + 4 – iz – 2i
z(1 – i) = – 4 + 2i
z = – 3 – i

3.) Consider the cubic function f(x) = ax3 + 2x2 + 3x + 4. Find the value of a in each of the following cases.

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

4.) Solve the equations

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

x(x+1)=6
x2+x-6=0
x=2

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

x(x+3)=10
x2+3x-10=0
x=2

c) log(x+18) – logx = 1

(x+18)/x= 1

5.) Solve the equation tan2x = 3 for -π≤ x ≤ π

tanx = ±√3
For tanx = √3
x=π/3, x= -2π/3

For tanx = -√3
x=-π/3, x= 2π/3

6.) Solve the following system of equations

X + 3y – 2z = -6
2x + y + 3z = 7
3x – y + z = 6
Back substitution gives x = 1, y = -1, z = 2.

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.

Mean = ⅙ (1+1+2+3+4+5) = 8/3
Variance= ⅙ (1+1+4+9+16+25) – 64/9 = 20/9

8.) Let f(z) = z2 – 8z + 20

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 ± yi
z= (8±4i)/ 2 = 4±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)2 + 4

9.) Given the function f(x) = x2 – 3bx + (c+2), determine the values of b and c such as f(1) = 0 and f’(3) = 0.

f(1) = 1 – 3b + c + 2 = 0
f’(x) = 2x – 3b
f’(3) = 6 – 3b = 0
b = 2
1 – 3(2) + c + 2 = 0
c =3

10.) If ln(2x-1), find d2y/dx2

dy/dx = 2/2x-1
d2y/dx2 = 2(2x-1)-2(2)
d2y/dx2 = -4/(2x-1)2

11.) Find ∫((3x2 – x + 2√x)/3√x)dx

= ∫(x1.5 – ⅓ x0.5 + ⅔ )dx = (x2.5/2.5) – (x1.5/4.5) + (⅔)x + c

Section B

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

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) = 6x2 + 6x – 12
Step 2: Set f'(x) = 0 to find stationary points
6x2 + 6x – 12 = 0
Step 3: Solve for x
Dividing both sides by 6, we get:
x2+ 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)22 – 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 – y1 = m(x – x1)
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:
∫f(x)dx = ∫(2x3+ 3x2 – 12x + 5)dx
Integrating each term separately:
∫2x3dx + ∫3x2 dx – ∫12xdx + ∫5dx
= (1/2)x4 + x3 – 6x2 + 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:
∫[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) = 6x2 + 6x – 12. Setting f'(x) = 0 and solving for x:
6x2 + 6x – 12 = 0
Dividing both sides by 6, we get:
x2 + 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) = 6x2 + 6x – 12. Taking the derivative of f'(x), we get:
f”(x) = (6x2 + 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).