Download Solution 1 - CUHK Mathematics

THE CHINESE UNIVERSITY OF HONG KONG
Department of Mathematics
MATH1020 General Mathematics (Fall 2016)
Coursework 1 Suggested Solution
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Please attempt to solve all the problems as exercises. This coursework will not be graded.
2
1. Answer the following questions:
(a) Assume that all radicands represent nonp-negative real numbers. Simplify each
expression by removing as many as factors as possible from under the radical.
√
p
a2 b
3
3
8
9
(ii)
(i) x 64x y z ;
.
ab−1
√
(b) Simplify m2 − 4m + 4, if possible. (Hint: Consider the cases m ≥ 2 and
m < 2 separately.)
(c) Assume all radicands are positive and denominators are non-zero. Simplify
each expression by rationalizing the denominator:
3
√ ;
1+ 2
√
s−3
(iii)
;
s−9
(i)
1
√ ;
√
√r − √3
s+ s+3
√
(iv) √
.
s− s+3
(ii)
(d) Find the domain of each function:
1
;
x+4
7x
(iii) f (x) = √
;
4−x
(i)
f (x) =
1
f (x) = √
;
3
x−1
x+4
.
(iv) f (x) = 2
x + 3x + 2
(ii)
3
p
(a)(i) x 3 64x3 y 8 z 9 = x(4xy 8/3 z 3 ) = 4x2 y 8/3 z 3 .
√
a2 b
(a)(ii) −1 = a2−1 b(1/2)−(−1) = ab3/2 .
ab
(
p
m−2
if m ≥ 2;
(b) m2 − 4m + 4 = (m − 2)2 = |m − 2| =
−m + 2 if m < 2.
√
√
√
3(1 − 2)
3−3 2
3
√ =
√
√ = 2
= −3 + 3 2.
(c)(i)
1 −2
1+ 2
(1 + 2)(1 − 2)
√
√
√
√
1
r+ 3
r+ 3
√ = √
√ √
√ =
(c)(ii) √
.
r−3
r− 3
( r − 3)( r + 3)
√
s−3
(c)(iii)
. The given denominator is already rationalized.
s−9
√
Remark. If the question asked for rationalization of numerator, the answer would be
√
√
1
s−3
s−3
√
=√
.
= √
s−9
( s − 3)( s + 3)
s+3
(c)(iv)
√
√
√
√
√
√
s+ s+3
( s + s + 3)2
( s + s + 3)2
√
√
√
= √
=
√
√
s − (s + 3)
s− s+3
( s − s + 3)( s + s + 3)
√
√
2
s + s + 3 + 2 s + 3s
2s + 3 + 2 s2 + 3s
=
=−
.
−3
3
4
(d)(i) (−∞, −4) ∪ (−4, ∞) or R \ {−4}
8
6
4
y
2
0
−2
−4
−6
−8
−8
−6
−4
−2
0
2
4
6
8
2
4
6
8
x
(d)(ii) (−∞, 1) ∪ (1, ∞) or R \ {1}
8
6
4
y
2
0
−2
−4
−6
−8
−8
−6
−4
−2
0
x
5
(d)(iii) (−∞, 4)
200
150
y
100
50
0
−50
−20
−15
−10
−5
0
5
x
(d)(iv) (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞) or R \ {−1, −2}, since x2 + 3x + 2 = (x + 1)(x + 2).
50
40
30
20
y
10
0
−10
−20
−30
−40
−50
−4
−3
−2
−1
x
0
1
2
6
2. Answer the following questions:
(a) Write each union as a single interval.
(i) (−∞, −3) ∪ (−5, ∞);
(ii) (−∞, −10] ∪ (−∞, −8].
(b) Write each intersection as a single interval.
(i) (−∞, −3) ∩ [−5, ∞);
(ii) [−20, −10] ∩ (−∞, −8].
(c) Write each set as an interval or as a union of two intervals.
1
;
(ii) {x| |2x − 3| ≥ 4} ;
(i)
x| |x − 4| < 10
2
(iii) {x| x + 4x > −3} .
(a)(i) (−∞, ∞) = R,
(a)(ii) (−∞, −8],
since −3 > −5.
since −8 > −10.
(b)(i) [−5, −3), since −3 > −5.
(b)(ii) [−20, −10], since −8 > −10.
(c)(i) (3.9, 4.1), since |x − 4| <
1
⇔ −0.1 < x − 4 < 0.1 ⇔ 3.9 < x < 4.1.
10
(c)(ii) (−∞, − 12 ] ∪ [ 72 , ∞), since
|2x − 3| ≥ 4 ⇔ 2x − 3 ≥ 4 or 2x − 3 ≤ −4 ⇔ x ≥
7
1
or x ≤ − .
2
2
(c)(iii) (−∞, −3) ∪ (−1, ∞), since
x2 + 4x > −3 ⇔ (x + 3)(x + 1) > 0 ⇔ x < −3 or x > −1.
7
3. Consider the function
(
1,
if x < 1;
f (x) =
2x − 3, if x ≥ 1.
(a) Sketch the graph of f (x) for −4 ≤ x ≤ 4.
(b) Evaluate and simplify each of the following expressions.
(i) f (1);
(iii) f (1 + ) where > 0;
(v) f (|x| + 2).
(ii) f (−2);
(iv) f (1 + ) where < 0;
(a)
6
5
4
y
3
2
1
0
−1
−2
−4
−3
−2
−1
0
1
2
3
4
x
(b)(i) f (1) = −1, since 2(1) − 3 = −1.
(b)(ii) f (−2) = 1, since −2 < 1.
(b)(iii) Since > 0, 1 + > 1, f (1 + ) = 2(1 + ) − 3 = 2 − 1.
(b)(iv) Since < 0, 1 + < 1, f (1 + ) = 1.
(b)(v) Since |x| + 2 > 1, f (|x| + 2) = 2(|x| + 2) − 3 = 2|x| + 1.
8
4. Recall that
(
x,
if x ≥ 0;
|x| =
−x, if x < 0.
Graph the following functions:
(a) f (x) = 3|2x + 6| − 4;
(b) f (x) = |x2 − 3x + 2|;
x−1
.
(c) f (x) =
|x + 1|
(Hint : Sketch the graph of y =
2
x−1
=1−
first.)
x+1
x+1
(a) Let g(x) = |x|. Then,
f (x) = 3g(2x + 6) − 4.
So, we can obtain the graph of f (x) by performing the following transformations on the
graph of g(x) :
(a) move it to the left by 6
(b) horizontally scale it by
1
2
(c) vertically scale it by 3
(d) move it down by 4
10
8
6
4
y
2
0
−2
−4
−6
−8
−10
−10
−8
−6
−4
−2
0
2
4
6
8
10
x
(Red indicates the part where the function is decreasing and blue indicates the part where
the function is increasing.)
9
(b) One can first sketch the graph of
y = x2 − 3x + 2 = (x − 1)(x − 2)
Then reflect the part that y < 0 (i.e. 1 < x < 2) about the x-axis.
10
8
6
4
y
2
0
−2
−4
−6
−8
−10
−10
−8
−6
−4
−2
0
x
2
4
6
8
10
10
(c)Let us first sketch the graph of
y = h(x) =
x−1
2
=1−
x+1
x+1
By letting g(x) = x1 , we have
h(x) = −2g(x + 1) + 1
Therefore, its graph can be obtained by performing the following transformations on the
graph of g(x) :
(a) move it to the left by 1
(b) reflect it about the x-axis
(c) vertically scale it by 2
(d) move it up by 1
The graph of f (x) can then be obtained by reflecting the part of h(x) that x < 1 about
the x-axis.
10
8
6
4
y
2
0
−2
−4
−6
−8
−10
−10
−8
−6
−4
−2
0
x
2
4
6
8
10
11
5. Consider the function f (x) = x2 +1. In this exercise, we will estimate the area under
the graph of f (x) for 0 ≤ x ≤ 3, that is, the area of the shaded region bounded by
the curve y = f (x) and the lines y = 0, x = 0 and x = 3 in Figure 1. This can be
done by approximating the region using rectangles.
Figure 1: Graph of f (x).
(a) In Figure 2(a), the horizontal interval [0,3] is divided into 3 equal segments. A
rectangle is drawn above each of these segments as shown. What is the total
area of the rectangles?
(b) In Figure 2(b), the horizontal interval [0,3] is divided into 6 equal segments
and 6 rectangles are drawn. What is the total area of these 6 rectangles?
(a) 3 rectangles
(b) 6 rectangles
Figure 2: Area Estimate by rectangles.
12
(c) Which answer in part (a) and (b) is a better estimate of the area under the
graph of f (x) for 0 ≤ x ≤ 3? Are the estimates greater or smaller than the
actual area under the graph of f (x)?
(d) (More difficult) Suppose the horizontal intervals [0,3] is divided into n equal
segments and n rectangles are drawn similarly as in part (a) and (b). Show
that the total area of these n rectangles is
9n(n − 1)(2n − 1)
+ 3.
2n3
The following identities may be useful:
n(n + 1)
1 + 2 + ... + n =
;
2
n(n + 1)(2n + 1)
12 + 22 + . . . + n2 =
.
6
(a.) The width of each rectangle is 1. The heights of the rectangles are
f (0) = 1, f (1) = 2, f (2) = 5.
Hence the total area is 1 · (1 + 2 + 5) = 8.
(b.) The width of each rectangle is 0.5. The heights of the rectangles are
f (0) = 1, f (0.5) = 1.25, f (1) = 2, f (1.5) = 3.25, f (2) = 5, f (2.5) = 7.25.
Hence the total area is 0.5 · (1 + 1.25 + 2 + 3.25 + 5 + 7.25) = 9.875.
(c.) Part (b) is a better estimation. The estimations are smaller than the actual area.
3
(d.) The width of each rectangle is .
n
The heights of the rectangles are
2
3i
9i2
3
f 0+ ·i =
+ 1 = 2 + 1,
n
n
n
where i = 0, 1, 2, ..., n − 1. Hence the total area is
!
n−1 2
n−1
n−1
X
X
X
3
9i
3
9
·
+1 = ·
i2 +
1
2
2
n i=0 n
n
n i=0
i=0
9(n − 1)(n)(2n − 1)
3
= ·
+n
n
6n2
9n(n − 1)(2n − 1)
=
+ 3.
2n3
13
6. Find the average rate of change of f over the interval [a, x]; that is
f (x) − f (a)
,
x−a
for each function. Be sure to simplify.
(a) f (x) = x2 − 2x + 3;
(b) f (x) = x3 ;
3
;
(c) f (x) =
1−x
√
(d) f (x) = x + 4.
(a.)
f (x) − f (a)
x2 − 2x + 3 − a2 + 2a − 3
(x − a)(x + a) − 2(x − a)
=
=
x−a
x−a
x−a
= x + a − 2.
(b.)
x 3 − a3
(x − a)(x2 + xa + a2 )
f (x) − f (a)
=
=
= x2 + xa + a2
x−a
x−a
x−a
(c.)
3(1−a)−3(1−x)
(1−x)(1−a)
3
− 1−a
=
x−a
3
=
(1 − a)(1 − x)
f (x) − f (a)
=
x−a
3
1−x
x−a
=
−3a + 3x
(1 − a)(1 − x)(x − a)
(d.)
√
√
√
√
√
x+4− a−4
x− a
x− a
√ √
√
=
= √
x−a
x−a
( x − a)( x + a)
1
√
=√
x+ a
f (x) − f (a)
=
x−a
√
14
7. Figure 3 shows the graphs of each of the six basic trigonometric functions.
1.5
1
1
y
0
0
−0.5
−1
−1
−1.5
−2pi
−3pi/2
−pi
−pi/2
0
pi/2
pi
3pi/2
−5pi/2
2pi
−2pi
−3pi/2
−pi
y
(a) y = sin (x)
5
4
4
3
3
2
2
1
1
y
0
−1
−2
−2
−3
−3
−4
−4
−pi/2
0
pi/2
pi
−5
−2pi
3pi/2
−3pi/2
−pi
−pi/2
(c) y = tan (x)
5
4
4
3
3
2
2
1
1
y
0
−1
−2
−2
−3
−3
−4
−4
0
2pi
5pi/2
0
pi/2
pi
3pi/2
2pi
0
−1
−pi/2
3pi/2
(d) y = cot (x) = 1/ tan (x)
5
−pi
pi
x (radians)
x (radians)
−5
−3pi/2
pi/2
0
−1
−pi
0
(b) y = cos (x)
5
−5
−3pi/2
−pi/2
x (radians)
x (radians)
y
y
0.5
pi/2
pi
x (radians)
(e) y = sec (x) = 1/ cos (x)
3pi/2
−5
−2pi
−3pi/2
−pi
−pi/2
0
pi/2
pi
3pi/2
2pi
x (radians)
(f) y = cosec (x) = 1/ sin (x)
Figure 3: Six trigonometric functions with restricted domains.
The period of each trigonometric function is the length of each cycle, that is, the
period is the horizontal distance between any point on the curve and the next
corresponding point in the next cycle where the graph starts to repeat itself.
The amplitude of the sine and cosine function is one-half the distance difference
between the maximum and minimum values of the function. Here are graphing
facts and relationships for the six trigonometric functions.
15

y




y
2π
:
period =

b
y



y
= a sin (bx + c)
amplitude = |a|
= a cos (bx + c)
amplitude = |a|
= a sec (bx + c)
= a cosec (bx + c)
(
y = a tan (bx + c)
π
period = :
b
y = a cot (bx + c)
Answer the following questions:
(a) Graph y = −2 sin (12x) by hand;
x
1
by hand;
(b) Graph y = cos
3
4
π
(c) Graph y = − sin 3x −
by hand;
2
π
by hand.
(d) Graph y = tan 3x −
2
2π
2π
π
(a.) The amplitude is 2 and the period is p =
=
= . The effect of the negative
b
12
6
sign is to reflect the curve about x-axis.
3
2
y
1
0
−1
−2
−3
−pi
−3pi/4
−pi/2
−pi/4
0
x
pi/4
pi/2
3pi/4
pi
16
(b) The amplitude is
2π
2π
1
and the period is
=
= 8π.
3
b
1/4
1.5
1
y
0.5
0
−0.5
−1
−1.5
−4pi
−2pi
0
2pi
4pi
x
2π
c
−π/2
π
2π
=
. The phase shift is =
=− ,
(c) The amplitude is 1 and the period is
b
3
b
3
6
π
i.e.
to the right. The effect of the negative sign is to reflect the curve about the x-axis.
6
1.5
1
y
0.5
0
−0.5
−1
−1.5
−2pi
0
x
2pi
17
(d) The period is
π
π
c
−π/2
π
π
= and the phase shift is =
= − , i.e.
to the right.
b
3
b
3
6
6
The asymptotes are
x=
1
π
π π (n + 1)π
1
nπ + − c =
nπ + +
=
b
2
3
2
2
3
we can simply write as x =
nπ
, n ∈ Z.
3
20
15
10
y
5
0
−5
−10
−15
−20
−pi/2
−pi/3
−pi/6
0
x
pi/6
pi/3
pi/2
18
8. (Optional) Let us study the so-called cardinal sine function. The standard sinc x
function without normalization is defined by

 sin x
x 6= 0;
sinc x =
x
1
x = 0,
as illustrated in Figure 4.
1.5
1
y
0.5
0
−0.5
−1
−1.5
−5pi
−4pi
−3pi
−2pi
−pi
0
pi
2pi
3pi
4pi
5pi
x
Figure 4: The standard sinc function without normization.
Find the zero crossing points of the normalized sinc function:

 sin πx
x 6= 0;
sinc x =
πx
1
x = 0.
Inspection of Figure 4 shows that the sinc function which has not been normalized has
the same zero crossing points as sin x, with exception of the crossing x = 0, that is at
x = ±π, ±2π, ±3π, · · · , ±nπ, · · · .
The normalized sinc function is 0 when sin(πx) = 0 except for x = 0. Hence,
sin(x) = 0
when x = · · · , −3π, −2π, −1π, 0, 1π, 2π, 3π, · · ·
and so
sin(πx) = 0
when x = · · · , −3, −2, −1, 0, 1, 2, 3, · · · .
19
Noting that sinc(0) is defined to have a value of 1, then we see that
sinc(x) = 0
when x = · · · , −3, −2, −1, 1, 2, 3, · · ·
or
sinc(x) = 0
when x = ±n for n = 1, 2, 3, · · ·
The plot of the normalized sinc function is shown below.
1.5
y
1
0.5
0
−0.5
−15
−10
−5
0
x
5
10
15