辅导案例-MATH1131

欢迎使用51辅导,51作业君孵化低价透明的学长辅导平台,服务保持优质,平均费用压低50%以上! 51fudao.top
THE UNIVERSITY OF
NEW SOUTH WALES
SCHOOL OF
MATHEMATICS AND
STATISTICS
Term 2 2019
MATH1131
MATHEMATICS 1A
*. TIME ALLOWED – 2 hours
9. TOTAL NUMBER OF QUESTIONS – 3
B. ANSWER ALL QUESTIONS
C. THE QUESTIONS ARE OF EQUAL VALUE
F. ANSWER EACH QUESTION IN A
SEPARATE BOOK
J. THIS PAPER MAY BE RETAINED BY THE
CANDIDATE
L. ONLY CALCULATORS WITH AN AFFIXED
ʼnUNSW APPROVEDŊ STICKER MAY BE
USED
N. A SHORT TABLE OF INTEGRALS IS
APPENDED TO THE PAPER
P. TO OBTAIN FULL MARKS YOUR
ANSWERS MUST NOT ONLY BE
CORRECT BUT ALSO ADEQUATELY
EXPLAINED CLEARLY WRIT- TEN AND
LOGICALLY SET OUT.
All answers must be written in ink. Except
where they are expressly required pencils may
only be used for drawing sketching or
graphical work.
*. i Use integration by parts to evaluate ∫ 0
√3 tan−1x dx.
ii. a Write down the definitions of
sinhx and coshx in terms of the
exponential function. b Evaluate
∫ 0 ln3 coshx + 1 sinhx dx.
iii. Let u = ⎜⎜⎜⎜⎝ ⎛ −1 −2 3 5 1 ⎟⎟⎟⎟⎠ ⎞ and v =
⎜⎜⎜⎜⎝ ⎛ −3 −6 β 9 3 ⎟⎟⎟⎟⎠ ⎞ be two vectors in
R5.
a. Find β if u and v are parallel.
b. Find β if u and v are perpendicular.
iv. Let a = ⎛⎝ 41 6 ⎞⎠ and b = ⎛⎝ −20 10 ⎞⎠ be
two vectors in R3.
a. Calculate Projba the projection of
a onto b.
b. Hence find a vector c not equal to
a in the plane spanned by a and b
such that |a| = |c| and Projba =
Projbc.
v. For the following system of equations
use Gaussian elimination to deter- mine
which values of λ if any will yield:
a. no solution
b. infinite solutions
c. a unique solution.
x. y + z = 4
x. λy + 2z = 5
2x+ λ+ 1y + λ2 − 1z = λ+ 7.
vi. Let fx = 2 + 1 x and consider the uniform
partition Pn of the interval
r. 1@ given by
Pn = { 0 n 1 n 2 . . . n− n 1 n n } .
a. Sketch a graph of f for 0 ≤ x ≤ 1 and
indicate the rectangles defining the lower
Riemann sum for f on P4.
b. Write down an expression for the lower
Riemann sum for f on Pn.
c. Hence or otherwise show that
n→∞ lim 2n+ 1 1 + 2n+ 1 2 + z z z+ 3n 1 = ln
3− ln 2.
2. i Let the region S be defined as
S = {z Ʉ C : |z − 2− 2i| ≤ 2 and Rez > 1}.
a. Sketch the region S on a carefully labelled
Argand diagram.
b. State in a+ ib form the complex number
in S of maximum modulus.
ii. A table sits on horizontal ground. In
Figure 1 a bird stands on the table
directly above a cat sitting on the ground
and the distance between the tops of
their heads is 105 cm as shown. In Figure
2 the cat and the bird have swapped
places and the distance between the
tops of their heads is now 145 cm.
x
Figure 1
105 cm
y
145 cm
z
Figure 2
Let x be the height of the bird y be the height
of the cat and z be the height of the table with
all measurements made in cm.
a. It follows from Figure 1 that x− y+ z = 105.
By considering Figure
9. write down a second linear equation
in x y and z.
b. By reducing an appropriate augmented
matrix to echelon form and back-
substituting find the height z of the table
and express the heights of the bird and
cat in terms of a parameter.
c. Given that the sum of the heights of the
bird and cat is 112 cm find the height of
the bird.
iii Consider the following pentagon ABCDE
with five equal-length sides.
C
B
u
A
E
Let u = −→AB and v = −−→ED. Observe that −
−→BD is parallel to −→AE and hence
D
v
BD −−→ = λ −→AE for some λ > 1.
Similarly −→AC = λ ED −−→ = λv and −−→EC
= λ −→AB = λu.
a. By considering the triangle ACE prove
that −→AE = λ v − u
b. Find another expression for −→AE in
terms of λ v and u and hence find the
exact value of λ expressing your answer
in surd form.
iv Use the following Maple output to explain
why A is invertible and to calculate A−1 for the
given matrix A. Give reasons for your answer.
> withLinearAlgebra: > A := <<2/3-1/3-2/3>|
<-1/32/3-2/3>|<-2/3-2/3-1/3>>
> DeterminantA
> A^3-A
A := ⎢⎢⎢⎢⎢⎢⎣ ⎡ − − 3 2 3 3 2 1 − − 3 2 3 3 2 1 − −
− 3 3 3 2 2 1 ⎥⎥⎥⎥⎥⎥⎦ ⎤ −1 ⎡ ⎣0 0 00 0 0 0 0 0 ⎦ ⎤
v a By considering seventh roots of unity find
a non-real solution to the
equation
1− 1 + w w 7 = 1.
b Show that your solution in part a is purely
imaginary.
USE A SEPARATE BOOK
CLEARLY MARKED
QUESTION 3
3. i Suppose that y = hx is a continuous
function over the real line with
the property that
lim hx = 2.
x→∞
a. Draw a possible sketch of the graph of h.
b. Using the formal definition of the limit
explain why there exists a real number M
with the property that
x > M ⇒ hx > 1.
c. Hence prove that ∫ 0 ∞ hx dx is a
divergent improper integral.
ii. The function f : R→ R is defined by
fx = ex x .
a. Find and classify the stationary point of f .
b. Using LŇHopitalŇs rule evaluate lim fx.
x→∞
c. What is the range of f?
d. Sketch the graph of f .
e. Define a function g : >1∞→ 0 e−1@ by
the rule
gx = fx
Explain why the function g is invertible while f is
not. f Let g−1 be the inverse function of g.
Show that
g−1x = xeg−1x
g. On what interval is g−1 differentiable?
h. Show that the derivative of the inverse of
g satisfies
>g−1x@′ = x1− g −1x g−1x .
iii. Suppose that both the functions f : R→ R
and g : R→ R are continuous
n the closed interval >a b@ and
differentiable on the open interval
a b with ga 6= gb and g′c 6=
0. Let
hx = fb− fagx− gb− gafx. By
applying the Mean Value theorem to h prove
that there exists c Ʉ a b such that
f ′c
fb− fa
=
g′c gb− ga .
C. INTEGRALS∫ |k|∫ eax x 1 dx dx = = ln a 1
|x|+ eax + C BASIC = ln |kx| C = ln
D. C∫ C∫ 1∫ sec2 sin cos ax dx ax ax ax = dx
dx dx 1ln = = = aa −1 a 1 a 1 x a sin tan +
cos ax+ C ax+ ax+ a 6=
E. C∫ C∫ C∫ C∫ tan sec sinh cot cosh
cosec2ax ax ax ax ax ax dx dx dx dx dx =
= dx = = = a a 1 1 a 1 = a 1 a 1 ln ln ln cosh
sinh −1 | | | sec sin a sec cot ax+ ax+ ax|+
ax+ ax|+ ax+ tan ax|+
F. C∫ C∫ a2 sech2ax cosech2ax dx + x2 =
dx dx a 1 = tan−1 = a 1 tanh −1 a x a coth
+ ax+ ax+
G. a2 dx − x2 = a 1 tanh−1 x a + C
x. < a
= 1 coth−1 x + C
a a
|x| > a > 0
a2∫ C∫ C∫ dx√ dx√ dx√ x2 x2 a2 − + − x2 a2
a2 = = = = 2a 1 sinh−1 sin−1 cosh−1 ln ɈɈɈɈ a−
a+ x a x a x a + x x + + ɈɈɈɈ+ C C x x > 2 a 6= >
0
END OF
EXAMINATION
2019/10/14 15)14
第 1 ⻚页(共 1 ⻚页)
51作业君

Email:51zuoyejun

@gmail.com

添加客服微信: abby12468