代做MATH1061: Mathematics 1A Semester 2, 2024 Assignment 2代写C/C++程序
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Semester 2, 2024
Assignment 2
Please justify your answers. Correct answers without adequate justification will not receive full marks. A plot on its own is not considered adequate justification.
1. Consider f(x) = 2x3 − 3x2 − 12x + 6.
(a) Find and classify the critical point(s) for f(x) on [−3; 3].
(b) Find the global maximum and minimum values of f(x) on [−3; 3].
(c) Does f(x) have any point of inflection on [−3; 3]? If it does, locate this point of inflection; if not, explain why.
2. (a) For integer n ≥ 1, use upper and lower Riemann sums with n equal subdivisions to find an upper and lower bound for the value of
You may use the following fact without proof:
(b) Evaluate dx using the definition of Riemann integral.
3. Let
(a) Find the 2nd order Taylor polynomial P2 (x) of f(x) centred at x = 0.
(b) Use the Lagrange form. of the remainder to obtain an upper bound for the remain- der R2 (x) when x = 1.
4. Let ℓ 1 and ℓ2 be two lines in space defined by the parametric equations:
ℓ 1 : x = −7 + 2s, y = 2 − 3s, z = 1 − s (s ∈ R)
ℓ2 : x = −12 + 3t, y = 2 − 2t, z = −13 + 4t (t ∈ R)
Let P be the plane that contains ℓ 1 and ℓ2 .
(a) Find the point of intersection of ℓ 1 and ℓ2 . (b) Find a general equation for P.
5. Let λ ∈ R, and consider the system of linear equations in the variables x,y,z given by
x − z = −2
2x + λy + 3z = 0 3
x + 2y + λz = −2
(a) Row reduce the corresponding augmented matrix to row echelon form.
(b) Find the values of the constant λ for which the system has
(i) no solutions
(ii) exactly one solution
(iii) infinitely many solutions
6. In a colony of 5200 bats, the maximum age of a bat is three years. We divide the female bats into three age groups of equal duration. We say that a female bat
• is in Age Group 1 if her age is between 0 and 1 years;
• is in Age Group 2 if her age is between 1 and 2 years;
• is in Age Group 3 if her age is between 2 and 3 years.
We have the following data about birth rates in each age group:
• bats in Age Group 1 have an average of 0.25 daughters per female;
• bats in Age Group 2 have an average of 6 daughters per female;
• bats in Age Group 3 have an average of 5 daughters per female.
We have the following data about survival rates in each age group:
• 50% of bats in Age Group 1 survive to Age Group 2;
• 40% of bats in Age Group 2 survive to Age Group 3.
We encode the birth rates and survival rates in a matrix L where
• bi is the average number of daughters per female in Age Group i; and
• si is the proportion of Age Group i that survives to Age Group i + 1.
The matrix L is
where the middle column has been filled in for you.
(a) Finish writing down the matrix L.
For year n we define the vector where
• xn is the number of female bats in Age Group 1;
• yn is the number of female bats in Age Group 2; and
• zn is the number of female bats in Age Group 3.
This means that for year n + 1 we have
xn+1 = Lxn.
Suppose initially we have 1600 bats in Age Group 1, 1600 bats in Age Group 2, and 2000 bats in Age Group 3; or in other words,
Find the number of bats in each age group in year two, i.e. find x2 .
(c) By considering the system
(L − 2I3)x = 0,
find all the vectors x that satisfy Lx = 2x.
(d) Suppose that instead of the initial conditions in part (b), we initially have 4000 bats in Age Group 1, 1000 in Age Group 2, and 200 in Age Group 3. Find the number of bats in each age group in year n = 5.