vectors randomly

advertisement
HL Mathematics Summer Packet
Statistics
3
1
4
4
1. a. Expand ( + )5
b. In New Zealand in 1946 there were two different coins of value two
shellings. These were ‘normal’ kiwis and ‘flat back’ kiwis, in the ration 3:1.
From a very large batch of 1946 two shelling coins, five were selected at
random. Find the probability that:
i. two were ‘flat backs’
ii. At least 3 were ‘flat backs’
iii. at most were ‘normal’ kiwis
2. An infectious flu virus is spreading through a school. The probability of
a randomly selected student having the flu next week is 0.3.
a. Mr C has a class of 25 students,
i. Calculate the probability that w or more students will have the flu
next week.
ii. If more than 20% of the students have the flu next week, a class
test will have to be cancelled. What is the probability that the test will be
cancelled.
b. If the school has 350 students, find the expected number that will
have the flu next week.
3. Apples from a grower’s crop were normally distributed with a mean 173
grams and standard deviation 34 grams. Apples weighing less than 130
grams were too small to sell.
a. Find the proportion of apples from this crop which were too small to
sell.
b. find the probability that in a picker’s basket of 100 apples, up to 10
apples were too small to sell
4. A university professor determines that no more than 80% of this year’s
History candidates should pass the final examination. The examination
results were approximately normally distributed with a mean 62 and
standard deviation 13find the lowest schore necessary to pass the
examination.
5. The length of steel rods produced by a machine is normally distributed
with a standard deviation of 3 mm. It is found that 2% of all rods are less
than 25 mm long. Find the mean length of rods produced by the machine.
Logarithms and Exponential functions
Simplify:
1. 𝑒 4𝑙𝑛π‘₯
2. 𝑙𝑛𝑒 5
4. 10π‘™π‘œπ‘”π‘₯+π‘™π‘œπ‘”3
5. ln⁑( π‘₯ )
3.ln⁑(√𝑒)
1
6.
𝑒
π‘™π‘œπ‘”π‘₯ 2
π‘™π‘œπ‘”3 9
Solve for x
7. π‘™π‘œπ‘”π‘₯ = 3
8. π‘™π‘œπ‘”3 (π‘₯ + 2) = 1.732
Solve exactly for x
10. 𝑒 2π‘₯ = 3𝑒 π‘₯
π‘₯
9. π‘™π‘œπ‘”2 ( ) = −0.671
10
11. 𝑒 2π‘₯ − 7𝑒 π‘₯ + 12 = 0
Complex Numbers
1. If 𝑧 = 5 − 2𝑖 and 𝑀 = 2 + 𝑖 Find in simplest rectangular form (a + bi)
a. 𝑧 + 𝑀
b. 2𝑧
c. 𝑖𝑀
d. 2𝑧 = 3𝑀
e. 𝑧𝑀
f.
2. Simplify:
b.
1
2−𝑖
−
a.
𝑧
𝑀
𝑖
1−21
2
2+𝑖
3. Given Re means the real part and Im means the imaginary part of a
complex number. If 𝑧 = 2 + 𝑖 and 𝑀 = ⁑ −1 + 2𝑖 find:
a. Im(4z-3w)
c.Im(iz2)
b. Re(zw)
4. For polar form, a complex number is 𝑧 = π‘Ÿ(π‘π‘œπ‘ πœƒ + π‘–π‘ π‘–π‘›πœƒ) or 𝑧 = π‘Ÿπ‘π‘–π‘ (πœƒ).
Where r is the magnitude and  is arctan
Find the polar form of a) 2 + 6𝑖
b) 0
𝑏
π‘Ž
(much like a vector)
c) −1 − 𝑖
d) −𝑖
Trigonometric Identities
Prove each identity by transforming the left side into one or the expressions
listed on the right.
SHOW ALL YOUR WORK
1.
2.
π‘π‘œπ‘ πœƒ
1−π‘ π‘–π‘›πœƒ
+
1−π‘ π‘–π‘›πœ—
π‘π‘œπ‘ πœƒ
𝑐𝑠𝑐 3 πœƒ−π‘π‘ π‘πœƒπ‘π‘œπ‘‘ 2 πœ—
π‘π‘ π‘πœƒ
3. 1 −
π‘π‘œπ‘‘ 2 πœƒ+π‘π‘œπ‘‘πœƒ
π‘π‘œπ‘‘ 2 πœƒ−1
=
a. 2π‘ π‘’π‘πœƒ
b. π‘‘π‘Žπ‘›πœƒ
=
a. π‘ π‘–π‘›πœƒ
b. 1
+3=
a. 3π‘π‘œπ‘‘πœƒ − 4
b.
4.
𝑠𝑖𝑛3 πœƒ−π‘π‘œπ‘  3 πœƒ
1−𝑠𝑖𝑛2 πœƒπ‘π‘œπ‘  2 πœƒ
=
a.
3π‘π‘œπ‘ πœƒ−4π‘ π‘–π‘›πœƒ
π‘π‘œπ‘ πœƒ−π‘ π‘–π‘›πœƒ
π‘ π‘–π‘›πœƒ−π‘π‘œπ‘ πœƒ
1−π‘ π‘–π‘›πœƒπ‘π‘œπ‘ πœƒ
b.1 + π‘ π‘–π‘›πœƒπ‘π‘œπ‘ πœƒ
5. π‘ π‘’π‘πœƒ + 1 +
1−π‘‘π‘Žπ‘›2 πœƒ
π‘ π‘’π‘πœƒ−1
=
a.
π‘π‘œπ‘ πœƒ
1−π‘π‘œπ‘ πœƒ
b. π‘π‘œπ‘ πœƒ + 1
6. Show that (π‘π‘œπ‘ πœƒ + π‘ π‘–π‘›πœƒ)2 = 1 + 𝑠𝑖𝑛2πœƒ
7. Solve for π‘₯ where 0 ≤ π‘₯ ≤ 2πœ‹
a). 𝑠𝑖𝑛2π‘₯ + 𝑠𝑖𝑛π‘₯ = 0
c) 2π‘π‘œπ‘  2 π‘₯ + π‘π‘œπ‘ π‘₯ − 1 = 0
b) π‘π‘œπ‘ 2π‘₯ + 3π‘π‘œπ‘ π‘₯ = 1
8. Use the double angle formula to show that
1
1
1
1
a)𝑠𝑖𝑛2 πœƒ = − π‘π‘œπ‘ 2πœƒ
b) π‘π‘œπ‘  2 πœƒ = + π‘π‘œπ‘ 2πœƒ
2
2
2
2
Derivatives: Find the derivative of the following functions
1. 𝑦 = 2𝑑 + π‘π‘œπ‘ 2𝑑⁑
3. 𝑔(π‘₯) =
2. 𝑓(π‘₯) = ln⁑(3π‘₯ − 5)
2π‘₯ 3 +3
4. 𝑦 = 3π‘₯𝑠𝑖𝑛(π‘₯ 2 + 7)
π‘₯−5
2
5. 𝑓(π‘₯) = 𝑒 π‘₯ + π‘₯ 3 − 2π‘₯ + 4
Find
6. β„Ž(𝑑) = 2𝑑(𝑑 3 + 2𝑑 − 7)3
𝑑2𝑦
𝑑π‘₯ 2
7. 𝑦 = π‘₯ − π‘₯ 3
9. If 𝑦 = sin(2π‘₯ + 3) Show that
8. 𝑦 = π‘₯𝑙𝑛π‘₯
𝑑2𝑦
𝑑π‘₯ 2
10. If 𝑦 = 2𝑠𝑖𝑛π‘₯ + 3π‘π‘œπ‘ π‘₯ Show that
+ 4𝑦 = 0
𝑑2𝑦
𝑑π‘₯ 2
+ 𝑦 = 0.
Integration
1. A gradient function is given by
dy
ο€½ 10e 2 x ο€­ 5 .
dx
When x = 0, y = 8. Find
the value of y when
x = 1.
2. ∫ 4π‘₯ 3 + 7π‘₯ − 2𝑑π‘₯
2
4. ∫1 (3π‘₯ 2 − 2)𝑑π‘₯
2𝑧𝑑𝑧
6. ∫ 3
√𝑧 2 +1
1
∫ 2π‘₯+3 𝑑π‘₯
3.
5. ∫
7.
𝑙𝑛π‘₯
𝑑𝑠
𝑑𝑑
π‘₯
𝑑π‘₯
πœ‹
= −4sin⁑(2𝑑 − ) s(0)=100
2
APPLICATIONS OF INTEGRATION AND DIFFERENTIATION
1. It is given that

3
1
(a) Write down
f (x)dx = 5.

3
1
2 f (x)dx.
(b) Find the value of

3
1
(3x2 + f (x))dx.
2. A part of the graph of y = 2x – x2 is given in the diagram below.
x-axis.
(a)
Write down an expression for this volume of
revolution.
(b)
Calculate this volume.
3. You are designing a poster to contain 50 in2 of printing with margins
of 4 in each at the top and bottom and 2 in at each side. What
overall dimensions will minimize the amount of paper used?
4. A car starts by moving from a fixed point A. Its velocity, v m s–1
after t seconds is given by v = 4t + 5 – 5e–t. Let d be the
displacement from A when t = 4.
(a) Write down an integral which represents d.
(b) Calculate the value of d.
VECTORS
1. Find the vector equation of the line that passes through
a) A(1,2,1) and B(-1,3,2)
b) C(0,1,2) and D(3,1,-1)
2. Find the parametric equation of the line which passes through (0,1,2) in
the direction of i + j – 2k.
3. A particle at P(x(t),y(t)) moves such that x(t)=1+2t and y(t) =2-5t with
𝑑 ≥ 0. The distances are in cm and the time is in seconds.
a) Find the initial position of P
b) Illustrate the initial part of the motion of P where t=0,1,2,3
c) Find the velocity vector of P
d) Find the speed of P
Download