MAT9004 Practice Exam Assignment PDF

MAT9004 Practice Exam
MAT9004 Practice Exam
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MAT9004 Practice Exam
1. f ( x )=−2 x3−9 x2−12 x+ 2
a. f 1 ( x) is the derivative of f (x) with respect to x
using thepower rule, subtraction and additional rule we obtain
f 1 ( x ) =−6 x2−18 x−12
For x ∈[−2,3]
b. f 1 ( x ) =−6 x2−18 x−12
f 11(x ) is the derivative of f 1 ( x ) with respect to x
Using the power rule, addition and the subtraction rule we obtain
f 11 ( x )=−12 x
Since x ∈[−2,3] then
f 11 ( x )=−12 x for x ∈[−2,3]
c. The stationery points are at point f 1 ( x )=0
Hence the stationery points will be at the roots of −6 x2−18 x−12
Using the quadratic equation x=−b ± √ b2−4 ac
2 a
We obtain the roots of the function at the points x=−1∧−2
The value of f ( x ) at the stationery points will be
f ( −1 ) =7∧f ( −2 ) =6
d. Finding local minimum and local maximum
The stationery points are at (−1,7 )∧(−2,6)
Now we use the sign test to determine if the points are maximum or minimum
x -2 -1 0
f 1 ( x) 0 -12
sign +ve -ve
Using -2 and 0 to do the sign test we can see that the sign changes from positive
to negative. When the sign changes from positive to negative, then this indicates
that we have a local maximum. Hence point (-1,7) is a local maximum point.
Now using the test sign, we test point (-2,6) using the values -2 and 0
x -3 -2 -1

MAT9004 Practice Exam
f 1 ( x) -2 0
sign -ve +ve
The sign is changing from negative to positive hence the point (-2,6) is a local
minimum.
e. Graphing the function f gives
From this graph we can observe two points which can be classified as global
maximum and minimum. That is point; (-1,7) and (-2,6)
2. M =[1 6
1 0]
a. Eigenvalues of M
Using the characteristic polynomial
[ 1−λ 6
1 0−λ ] =λ2 −λ−6= ( λ+2 )∗(λ−3)
obtaining the roots of the equation givesthe eigenvalues as
λ=−2
And
λ=3
b. Finding eigen vector
For every λ we find its own vectors
A−λ−E=(3 6
1 2)

MAT9004 Practice Exam
A−λE=0, so, we have a homogeneous system of linear equations we solve it by
Gaussian Elimination
(3 6 0
1 2 0 )¿1/ 3
R1 /3 → R1
( 1 2 0
1 2 0 ) ↲∗(−1)
R2−1∗R1 → R2 (1 2
0 0)
{x1 +2 x2=0 (1)
Find the variables x1from the equation of the system (1)
x1=−2 x2
X =(−2 x 2
x 2 )
let x2 =1 , v1 =(−2
1 )
2 λ 2=3
A−λ∗E=(−2 6
1 −3)
A−λ∗E=0, so, we have a homogeneous system of linear equations, we solve it
by Gaussian Elimination to obtain
{x1−3 x2=0 (1)
let x2 =1, then v2=(3
1)
c. Diagonalize M
The diagonal matrix (the diagonal entries are the eigenvalues −λ1 , λ2
D=(−2 0
0 3)
The matrix with the eigenvectors ( v1 , v2 ¿ as its columns
P=(−2 3
0 3)
then P−1=(−1 /5 3/5
1 /5 2/5)
A=P−1∗D∗P

MAT9004 Practice Exam
(−1
5
3
5
1
5
2
5 )∗(−2 0
0 3 )∗(−2 3
0 3 )=(
−4
5
33
5
4
5
12
5
)
3. Calculate probability
f ( x )=x5 + x2 + x, (xϵ [ 0,1]
probability of {x> 1
2 }=∫
1/ 5
1
x5 + x2 + x= [ 1
6 + 1
3 + 1
2 ] − [ 1
384 + 1
2 + 1
8 ]= 319
384
4. The profit function is
p ( a , b )=a b2−a2 +b2 +6 b−9 a+10
The optimal points are at the derivative of the profit function equals to zero.
The derivative with respect to b gives
¿ 2 ab+2 b+6=0
¿
The derivative with respect to a
b2−2 a−9=0
Now a= b2−9
2
Now 2∗b2−9
2 +2 b+6
¿ b2 +2 b−3=0
using thequadratic formulawe obtain b¿−3 amd 1
Since the value of b is minimum we take -3
This gives the value of a¿ b2−9
2 =0
5. Not divisible by 2 and not divisible by 5.
a. This means the last digit cannot be 0,2,4,6,8
Now we must pick 4 digits from a bucket of 10
This can be picked in the following way
The fist number can be picked in 9 ways
2nd in 10 ways
3rd in 10 ways
4th in 4 ways

MAT9004 Practice Exam
Hence the total prime like numbers between 1000 and 9999 is
9∗10∗10∗4=3600
b. Has at least one even digit
Here we select 4 numbers from a bucket of 10 in the following way
1st number 9 ways
2nd number 10 ways
3rd number 10ways
4th number 5 ways
Now let’s assume that the even digit is in the 1st number then the numbers will
be
4∗5∗5∗4=400
Leta now assume the even number is in the second digit
5∗5∗5∗4=500
Let’s assume the even number is in the 3rd digit
5∗5∗5∗4=500
The even digit cannot be in the fourth number as this will make the number
prime like anymore
In total the numbers are 400+ 500+500=1400
c. Prime like numbers not divisible by 3
Here we must choose the four numbers such that the sum of the digit is not
divisible by 3
1st digit 9 ways
2nd digit 10 ways
3rd digit 10 ways
4th digit 3 ways
The numbers are 9∗10∗10∗3=2700
6. Y uniformly distributed
a.
For u1
Y -1 0 1 2

MAT9004 Practice Exam
u1 2 0 0 2
Since Y is uniformly distributed all the values occur with equal probability the
expected value of expected value of u1= 2+0+0+2
4 =1
For u2
Y -1 0 1 2
u2 0 0.5 0 0.5
Since all the values occur with uniform probability the expected value will be
0+0.5+0+0.5
4 =0.25
b. Variance of u1=∑
u1
( u1−μ ) p(u1 )
( 2−1 )+ ( 0−1 ) + ( 0−1 ) + ( 2−1 ) = 0
4 =0
The variance is 0
Variance of u2
( 0−0.25 ) + ( 5−0.25 ) + ( 0−0.25 ) + ( 0.5−0.25 )= 0
4 =0
The variance is 0
c. Testing independence of u1 and u2
No, the equations are dependent as they share similar variance
7. A fair 6-sided dice is rolled 3 times
a. Probability that sum is 8
Possible outcomes are 63 =216
Each roll must have a value between 1 and 6
1+1+1=3 cannot be
2+2+2=6 cannot be
3+3+3=9 cannot be
The only way we can have a sum of 8 is ( 3,3,2 ) , ( 4,2,2 ) , ( 6,1,1 ) , ( 5,2,1 )
The roll is 3 times so there are 3∗4=12 possibilities
Thus, the probability is 12
216 =1/18
b. Probability of 1 given sum of 8

MAT9004 Practice Exam
¿ pre (∑ 8∧1 )
prob (∑ ¿ 8 ) =
3
216
1
18 =1/4
8.
a. Drawing graph
b. The adjacency matrix is given by A=
1 0 1
1 1 0
1 1 1
the graph does not have a spanning
tree.
c. No, there is no connection that can allow all eight students to seat in a way that
any of them knows both neighbours.
References
McQuarrie, D., 2003. Mathematical Methods for Scientists and Engineers, s.l.: University Science
Books.
Salas, S. L., Hille, E. & Etgen, G. J., 2007. Calculus: One and Several Variables. 10th ed. s.l.:Wiley.
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