# MAT 1375 Solutions for Sample Test 3

```MAT 1375 Solutions for Sample Test 3
(ii) x 2 K 5 x %K6
(i) 5 x K 2 O 10
Solution to (i): To solve the inequality 5 x K 2 O 10 we first must remove the absolute value. We can
replace the absolute value by writing G instead:
5 x K 2 O 10 / G 5 x K 2 O 10 /
xO
5 x K 2 O 10
K 5 x K 2 O 10
/
5 x O 12
/ xO
12
5
K5 x C 2 O 10 / K5 x O 8
12
8
or 5 x !K8 / x !K
5
5
8
5
The final answer in an interval notation is: x 2 KN,K
W
12
,CN
5
Solution to (ii): Draw the real line and mark the solutions of the quadratic equation
x 2 K 5 x C 6 = x K 2 x K 3 = 0, namely 2 and 3, on the line:
+
2
K
3
+
then, we can find the solution to the quadratic inequality
x 2 K 5 x C 6 % 0/ x K 2
xK3 % 0
by finding the sign of the algebraic expression x K 2
x K 3 in each of the intervals:
KN, 2 , 2, 3 , 3,CN
It is actually enough to find the sign in any one of these intervals and then alternate the signs for the
remaining intervals. It is easy to see the sign in the interval containing x = 0, then
0K2
0K3 = 6 O 0
so the sign of the quadratic expression x 2 K 5 x C 6 is positive in the interval KN, 2 , negative in 2, 3
and again positive in 3,CN (alternate the signs).
This is thanks to the fact that any polynomial is a continuous function (no jumps or gaps) and 2 and 3 are
roots (zeroes), so on both sides of the zeroes the signs must be different. Here is a plot of the polynomial
to see this better:
+
y = x2 K 5 x C 6 /
0
+
K
2
3
x
The final answer is: x 2 K 5 x C 6 % 0 / x 2 2, 3 .
2. (Additional Problem) Find the domain and the range of the functions. Graph them using your
calculator:
(i) f x =
400 x 2 K 1
(ii) g x = 2 x K 1 C 2
Solution (i): The domain we get by observing that what's inside the radical must be positive for the square
root to be defined:
400 x 2 K 1 O 0 / 20 x K 1
20 x C 1 O 0
The quadratic inequality above we can solve by looking at the quadratic equation:
20 x K 1
20 x C 1 = 0
and we readily obtain the roots because we already have a product of two factors being zero, so either one
or the other factor must be zero:
20 x K 1 = 0 or 20 x C 1 = 0 / x1 =
1
1
or x2 =K
20
20
1
1
1
, K ,
,
20
20 20
The two roots above specify 3 intervals on the real line: KN,K
We are interested in the intervals where our expression is positive: 20 x K 1
1
,CN
20
20 x C 1 O 0
To find the signs of our expression in each interval, it is enough to find the sign in one of these intervals
and then alternate the signs. For example, take x = 0, which is in the second interval and we see that our
expression is negative there: 20\$0 K 1 20\$0 C 1 =K1 ! 0 , so in the first interval as well as the 3rd
interval our expression must be positive, so the domain is then the first interval or the 3rd interval:
Domain f x
1
1
W
,CN
20
20
= KN,K
The range is where the y = f x values live but these values are given by a square root, so they all must be
positive or zero y = f x R 0 and this defines the range: Range f x = 0,CN
10
7.5
5
2.5
400\$x 2 K 1 /
y=
0
K0.4 K0.2
0.2
x
0.4
Solution (ii): Here, there is no restriction on the domain, so Domain g x = KN,CN . However,
the range is restricted because the absolute value is always ≥ 0 and when we add 2, then g x R 2 so
Range g x = 2,CN .
8
6
4
y = 2\$x K 1 C 2 /
2
K2 K1
0
1
2
3
x
3. (a) Find the sum of the first 65 terms in the arithmetic sequence: an : -6, -2, 2, 6, 10, ...
(b) Find the sum of the infinite geometric sequence: an : 24, -12, 6, -3, ...
Solution to (a): The common difference for the arithmetic sequence is
d = a2 K a1 =K2 K K6 =K2 C 6 = 4.
The sum of the first 65 term is:
65
>a
n=1
n
= 65 a1 C
65 65 K 1
65\$64
d = 65\$ K6 C
\$4 = 7930
2
2
a2
K12
1
1
1
=K . Since K =
! 1 the sum of the
a1
24
2
2
2
infinite geometric sequence is finite, that is a number we can compute:
Solution to (b): The common ratio is r =
N
>
n=1
an = 24 K 12 C 6 K 3 C\$\$\$ =
a1
1Kr
=
=
24
1
2
1K K
=
24
1C
1
2
=
24
24\$2
=
= 16
3
3
2
4. Find the equation of a line through the point K2, 4 and parallel to the line 5 x K 10 y = 7.
Solution: We can find the equation of the line through the given point if we know the slope of that line but
we know that it must be parallel to the line 5 x K 10 y = 7 and so it must have the same slope. The slope
we can find if we express y in terms of x:
5
7
1
7
xK
= xK
10
10
2
10
1
then the slope is the signed number in front of the variable, namely: m =
2
The line we are seeking must have the same slope. Now, given a point with coordinates x0 , y0 and a
slope m, we can write the equation of the line passing through this point having the given slope:
y = y0 C m x K x0 . Remember that y0 , m, x0 are given numbers and y, x are variables, representing the
1
coordinates of an arbitrary point on the line. The final answer we get by using x0 =K2, y0 = 4, m = :
2
1
1
1
1
y = 4C
x K K2 = 4 C x C 2 = 5 C x
2
2
2
2
5 x K 10 y = 7/ 5 x K 7 = 10 y / 10 y = 5 x K 7/ y =
5. Find the equation of the perpendicular bisector of the line segment joining the points
K2, 2 and 4,K2 .
K2 C 4 2 C K2
,
= 1, 0 .
2
2
Then, to find the equation of the line passing through the mid-point and perpendicular to the given line
segment, we also need the slope of this line. Let's denote this slope by m2 and let the slope of the line
through the two given points be m1 . Given that these two slopes represent two lines that are perpendicular,
1
we must have m1 .m2 =K1 , which gives us the slope we are seeking: m2 =K
. So, we just need to find
m1
the slope m1 of the line through the two given points:
Solution: We need to find first the mid-point of the line segment:
m1 =
y2 K y1
x2 K x1
=
K2 K 2
4 K K2
4
2
2
=K / m2 = K1 O K
6
3
3
=K
3
2
= K1 \$ K
=
3
2
Now, we use the Point-Slope form to write the equation of the line through the mid-point, perpendicular
to the line segment:
y = 0 C m2 x K 1 =
3
3
3
xK1 = xK
2
2
2
6. Let f x = 2 x 2 K 4 x K 1. Evaluate the expressions:
(i) f Kx
(ii) f 1 K x
(iii)
f xCh Kf x
h
Solution: (i) f Kx = 2 Kx
2
K 4 Kx K 1 = 2 x 2 C 4 x K 1
(ii)
f 1 K x = 2 1 K x 2 K 4 1 K x K 1 = 2 1 K 2 x C x 2 K 4 C 4 x K 1 =K4 x C 2 x 2 C 4 x K 3
f 1 K x = 2 x2 K 3
(iii)
f xCh Kf x
2 x C h 2 K 4 x C h K 1 K 2 x2 K 4 x K 1
=
h
h
2
2
2 x C 2 xh C h K 4 x K 4 h K 1 K 2 x 2 C 4 x C 1
=
=
h
=
2 x 2 C 4 xh C 2 h 2 K 4 x K 4 h K 1 K 2 x 2 C 4 x C 1
4 xh C 2 h 2 K 4 h
=
= 4 xK4C2 h
h
h
7. Find a cubic polynomial p x with roots of K2, 0, 0 and p 1 = 6. Express your answer as a
product of linear factors.
Solution: Since -2 and 0 are roots, p x must have the factors x K K2
but 0 must be a double root, so:
= x C 2 and x K 0 = x,
p x = a x C 2 x 2 , where a is a number we must find given the information p 1 = 6
p 1 = 6 = a 1 C 2 12 / 6 = 3 a / a = 2 / p x = 2 x C 2 x 2
8. Find all roots of the cubic polynomial f x = 2 x 3 K x 2 K 3 x K 1 exactly, using the rational root
test. Sketch a graph of f x .
Solution: The goal is to find the complete factorization of f x . By the rational root test, if there is a
m
rational root
then m =G 1, n =G 1, G 2, so all possibilities are:
n
m
1
1
1
=G 1, G
and if we check all these 4 possibilities, we see that K is a root, that is f K
= 0.
n
2
2
2
1
1
1
1
If K is a root, then x K K
= xC
must be a factor of f x = x C
\$g x and to find
2
2
2
2
g x we have to divide the polynomial f x by the linear factor:
2 x3 K x2 K 3 x K 1
g x =
=
= 2 x2 K 2 x K 2
1
1
xC
xC
2
2
f x
To find the complete factorization of g x = 2 x 2 K 2 x K 2, we need to find the roots with the quadratic
formula:
2 x 2 K 2 x K 2 = 0 / x1, 2 =
1G 5
2
/ g x = 2 x 2 K 2 x K 2 = x K x1
x K x2
Finally, we can write the complete factorization of the cubic polynomial:
1
2
f x = xC
1C 5
2
xK
xK
1K 5
2
1 1C 5 1K 5
which gives us all 3 roots: K ,
,
2
2
2
(the two negative roots are very close to each other around -0.5)
2
y = 2 x3 K x2 K 3 x K 1 /
K2
K1
0
K2
1
x
2
9. Identify if you have a circle or an ellipse and draw their graphs.
(i) x 2 K 4 x C y 2 C 2 y = 0
(ii) x 2 C 4 y 2 K 6 x K 8 y C 9 = 0
Solution (i): When we have x 2 C y 2 C\$\$\$ we know that we have a circle and to find the center and the
radius we just need to complete the squares:
4 2
4 2
2
x K4 xCy C2 y = x K4 xC
K
C y2 C 2 y C
2
2
2
2
2
2
2
2
2
2
K2 Cy C2 yC1 K1 = xK2 K2 C yC1 K1 = 0
2
/
2
xK2
2
2
C yC1
2
=5
The center of the circle is 2,K1 and the radius is
xK2
2
C yC1
2
=5/
5 :
2
K
2
2
2
= x 2 K 4 x C 22
2
x
y K2
4
Solution (ii): When we have a x 2 C b y 2 C\$\$\$ with a s b we know we have an ellipse but to find the
center of the ellipse we need to complete squares again:
6 2
6 2
x K6 xC4 y K8 yC9 = x K6 xC
K
C 4 y2 K 2 y C 9 =
2
2
2
2
2
2
2
2
x K 6 x C 3 K 3 C 4 y K 2 y C 1 K 1 C 9 = x K 3 2 K 32 C 4 y 2 K 2 y C 1 K 4 C 9 =
xK3 2
yK1 2
= xK3 2 C4 yK1 2 K4 = 0 /
C
=1
4
1
2
2
2
So, the center of the ellipse is 3, 1 and the major axis is the x-axis since 4 O 1. We can draw the ellipse
without having to find the foci:
2
xK3
4
2
C
yK1
1
2
=1/
y
1
0
1
2
3
x
4
5
10. The initial population of a colony is 10,000 and is decreasing exponentially at 1.5% rate per year.
• What is the size of the colony in 5 years?
• How long will it take for the population to be half of its initial amount?
Solution: The exponential decrease is given by the exponential function of time t
A t = P\$eKk\$t
where P = 10000 is the initial population, k = 1.5 % = 0.015 is the given rate, with minus because it is an
exp. decrease. The size of the colony in 5 years is:
A 5 = 10000\$eK0.015\$5 z 9277
Let's say that it will take x years for the population to be halved:
5000 = 10000\$eK0.015\$x / eK0.015\$x =
1
1
/ ln eK0.015\$x = ln
2
2
K0.015\$x = ln 2K1 =K1\$ln 2 =Kln 2 / x =
ln 2
= 46.21 years
0.015
11. For the functions below, find the amplitude, the period, the phase shift. Draw the graph over a oneperiod interval.
(i) y = 5 cos 4 x
2x
3
(ii) y =K2 sin
p
3
(iii) y =K3 sin 2 x K
Solution (i):
Amplitude = 5, period =
2p
p
=
, phase shift = 0
4
2
y = 5\$cos 4\$x /
K6
K4
K2
(ii): Ampl. = K2 = 2, period = 2 p O
2\$x
y =K2\$sin
3
2
4
6
x
2
2 p\$3
=
= 3 p, phase shift = 0
3
2
2
/
K6
K4
(iii) Ampl. = K3 = 3 , period =
phase shift is
6
2
K2
K6
K2
2
K1
4
6
x
2p
p
p
= p , to get the phase shift, set: 2 x K
=0/x=
so the
2
3
6
p
6
p
y =K3\$sin 2\$x K
3
3
/
K6
K4
K2
K2
2
4
x
6
```