IB Math AI SL Paper 1 Question Bank
1.) Let R be the region in the xy-plane that is enclosed by the graphs of y = x3 and y = x2 – 6x + 8.
a) Find the volume of the solid generated when R is rotated about the x-axis.
First, we need to find the limits of integration. From the equation of the upper bound, y = x3, we get x = y1/3.
From the equation of the lower bound, y = x2 – 6x + 8, we get x = 2 +√(y + 2) and x = 2 – √(y + 2).
So the volume of the solid generated when R is rotated about the x-axis is given by the definite integral:
∫[2-√(y+2), 2+√(y+2)] π(y2/3)2 dy
b) Find the volume of the solid generated when R is rotated about the y-axis.
First, we need to find the limits of integration. From the equation of the upper bound, y = x3, we get x = y1/3.
From the equation of the lower bound, y = x2 – 6x + 8, we get x = 2 + √(y + 2) and x = 2 – √(y + 2).
So the volume of the solid generated when R is rotated about the y-axis is given by the definite integral:
2π ∫[2-√(y+2), 2+√(y+2)] x(y1/3) dy
2.) 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
3.) Consider the quadratic -4x2+120x-800
a)
i) Find the roots.
x=10, x=20
ii) Hence express the quadratic in the form y= a(x-x1)(x-x2)
y= -4(x-10)(x-20)
b)
i) Find the coordinates of the vertex.
(15,100)
ii) Hence express the quadratic in the form y= a(x-h)2+k
y= -4(x-15)2+100
iii) Write down the equation of the axis of symmetry
x=15
iv) Write down the maximum value of y
Ymax= -100
c) Write down the y- intercept of the quadratic
y= -800
4.) Solve the equation: (ln x)2 – lnx2 +1 = 0
x2-2x+1=0
x= 1
lnx = 1
Hence, x = e
5.) Solve the equation 2sin2x = sin x for 0≤ x ≤ 2π
2sin2x – sinx = 0
sinx(2sinx – 1) = 0
Sinx = 0
and
2sinx -1 = 0
Sinx = 1/2
For sinx, x = 0, x= π, x = 2π
For sinx = 1/2, x=π/6, x= 5π/6
6.) Let f(x) = 2x2 – 12x +10. Find the tangent and normal line at x = 2.
Tangent line = y + 6 = -4(x -2)
Normal line = y + 6 = ¼ (x – 2)
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7.) Consider the following cuboid of dimensions 5*3*4, as shown.
a) Find the length AC.
AC2= 52+32
AC = √34
b) Find the length AD.
AD2 = √342+42 = √50
c) Find the angle of elevation from A to E.
tanEAB = 4/5
EAB = 38.7°
d) Find the angle of elevation from A to D.
tanDAC = 4/√34
DAC= 34.4°
8.) Solve 2sinx = tan x, where -π/2≤ x ≤ π/2
2sinxcosx – sinx = 0
sinx(2cosx – 1) = 0
Sinx = 0, cosx= 1/2
sinx=0, hence x= ±π/3
9.) In a class, 40 students take chemistry only, 30 take physics only, 20 take both chemistry and physics, and 60 take neither.
a) Find the probability
i) that a student takes physics given that the student takes chemistry.
P(P I C)= 20/(20+40) = ⅓
ii) that a student takes physics given that the student does not take chemistry.
P(P I C’’) = 30/ (30+60) = ⅓
b) State whether the events “taking chemistry” and “taking physics” aremutually exclusive, independent, or neither. Justify your answer.
P is independent of C since P(P I C) = P(P) = ⅓
10.) The random variable X is normally distributed with μ= 100. It given that P(X > 130) = 0.2 Write down the values of the following probabilities
a) P(X < 130)
P(X < 130) = 0.8
b) P(X < 70)
P(X < 70) = 0.2
c) P(100<X < 130)
P(100<X < 130) = 0.3
d) P(70<X < 130)
P(70<X < 130) = 0.6
11.) For the event A and B, P(A) = 0.6, P(B) = 0.8 and P(A∪B) = 1
a) Find P(A∩B)
P(A∩B)= 0.6 + 0.8 – 1 = 0.4
b) Find P(A’∪ B’)
P(A’∪ B’) = 0.
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