Download Fall 2001 Term Test One – Solutions

Okanagan University College
Math 112(71, 72 & 61), Fall 2001
Term Test One – Solutions
Instructor:
Clint Lee
010928/29
Student Name:
Total Marks:
35
Instructions. Do all parts of all 7 questions. Show all work and give explanations where required. You
may receive part marks for a question if your work is correct even if the final answer is incorrect. However,
if your answer is incorrect and no work or explanation is given, you will receive no marks. The number of
points for each question is given in the left margin, total 35.
[4]
1.
Find the equation of the line perpendicular to the line 3x − 2y = 4 that has its x-intercept at x = 3.
Remember that for two lines, neither of which is vertical, with slopes m1 and m2 , the product of the
slopes is m1 m2 = −1.
Find the slope of the given line by solving for y:
3x − 2y = 4 ⇒ 2y = 3x − 4 ⇒ y =
3
−2
2
2
3
so that the equation of the perpendicular line is m2 = − .
2
3
Since the perpendicular line has x-intercept at x = 3, a point on the line is (3, 0), thus, its equation is
Thus the slope of the given line is m1 =
2
2
2
y − 0 = − (x − 3) = − x + 2 ⇒ y = − x + 2
3
3
3
[4]
2.
Solve the inequality and express the solution as an interval: x2 − 4 ≤ 3x
Write the inequality as x2 − 3x − 4 ≤ 0. Then factor the quadratic x2 − 3x − 4 = (x − 4)(x + 1).
Thus, x2 − 3x − 4 = 0 for x = −1 or x = 4. These points divide the number line into three intervals:
(−∞, −1), (−1, 4), and (4, ∞). Check the sign of x2 − 3x − 4 at a test value in each interval. At x = −2
(in (−∞, −1)) , x2 − 3x − 4 is positive, at x = 0 (in (−1, 4)), x2 − 3x − 4 is negative, and at x = 5 (in
(4, ∞)), x2 − 3x − 4 is positive. Thus, x2 − 3x − 4 is negative only in the interval (−1, 4). Since we
allow equality the endpoints of the interval are included. Thus the solution is the interval [−1, 4].
[4]
3.
Use the facts that sin 2x = 2 sin x cos x, cos 2x = cos2 x − sin2 x, and cos2 x + sin2 x = 1 to prove the
identity
sin 4x = sin [2 (2x)]
= 2 sin 2x cos 2x
= 4 sin x cos x cos2 x − sin2 x
= 4 sin x cos x cos2 x − 1 − cos2 x
= 4 sin x cos x 2 cos2 x − 1
= 8 cos3 x sin x − 4 sin x cos x
Math 112(71, 72 & 61), Fall 2001
4.
Term Test One – Solutions
Page 2
A 10-cm length of wire is cut into two pieces. One piece of is bent into a square whose side length is x
and the other piece is formed into a circle of radius r. See the diagram below.
r
x
[2]
(a) Express the total area, A, of the square and the circle in terms of x and r. Remember that the
area of a circle is πr 2 .
A = x2 + πr2
[2]
(b) Use the fact that the total length of the wire is 10 centimetres to express the radius r in terms of
x. Remember that the circumference of a circle is 2πr.
4x + 2πr = 10 ⇒ 2πr = 10 − 4x ⇒ r =
[2]
5 − 2x
π
(c) Use your results in parts (a) and (b) to express the total area, A = f (x), enclosed by the square
and the circle as a funtion of the side length x of the square.
2
A = f (x) = x + π
5 − 2x
π
2
1
= x2 +
25 − 20x + 4x2
π
4
20
25
= 1+
x2 − x +
π
π
π
π+4
20
25
=
x2 − x +
π
π
π
Math 112(71, 72 & 61), Fall 2001
5.
Term Test One – Solutions
Page 3
Consider the piecewise defined function:

2

1 + 2x − x if − 1 ≤ x ≤ 1
f (x) = x2 − 2x + 4 if 1 < x ≤ 2


x+2
if 2 < x ≤ 4
The graph of f is shown below.
6
4
2
−2
[2]
2
4
(a) Give the domain and range of f , expressed as intervals.
Domain: [−1, 4]
Range: [−2, 2] ∪ (3, 6]
[1]
(b) Is f one-to-one? Explain.
Yes, f is one-to-one. It passes the horizontal line test. This means that there are no two different
x values for which f has the same value.
(c) Give the exact value of each of the following, or explain why the value does not exist:
[1]
i) f −1 (4) = 2 since f (2) = 4
[1]
ii) f −1 (3) does not exist, since there is not x value for which f (x) = 3
[2]
iii) f −1 (0) Hint: You have to solve a quadratic equation here.
To find f −1 (0) must solve the equation f (x) = 0 for x. From the graph we see the f (x) = 0
in the interval [−1, 1]. Thus the equation is
1 + 2x − x2 = 0 ⇒ x2 − 2x − 1 = 0
The solutions to this equation are
p
√
√
2 ± 4 − 4(1)(−1)
2± 8
=
=1± 2
x=
2
2
To get the solution in the interval [−1, 1] take the negative sign in the solutions above. Thus
√
f −1 (0) = 1 − 2 = −0.4142
Math 112(71, 72 & 61), Fall 2001
[5]
6.
Let f (x) =
Term Test One – Solutions
x
. Find and simplify
x+2
f (x + h) − f (x)
x+h
1
x
=
−
h
h x+h+2 x+2
1 (x + h)(x + 2) − x(x + h + 2)
=
h
(x + 2)(x + h + 2)
2
1 x + 2x + hx + 2h − x2 − xh − 2x
=
h
(x + 2)(x + h + 2)
2h
1
=
h (x + 2)(x + h + 2)
2
=
(x + 2)(x + h + 2)
[5]
7.
The graph of y = sin x is shown. Use appropriate
transformations to sketch the graph of
π x .
y = 2 + 3 sin
2
Specify each transformation as a vertical or horizontal shift, a vertical or horizontal stretch or
shrink, or a reflection across the x or y-axis. Further, give the amplitude and period of the resulting graph. You will not receive any marks for just
drawing the graph without any explanation.
The transformations are:
• Horizontal shrink by a factor of
2
= 0.6366
π
• Vertical stretch by a factor of 3
• Verticl shift up by 2 units
The the amplitude is A = 3 and period P is given by
π
P = 2π ⇒ P = 4
2
Page 4