Review/Tutorial for the Precalculus Placement Test
   
1. Find the radian measure of an angle whose degree measurement is 330o.
   
2. Which of the following numbers is the smallest?
 
a) b) c) d)
   
3. In a right triangle ABC, angle C is the right angle, side AC = 5
  and sinB = 0.64. Find the length of side AB to the nearest tenth.
 
   
4.
Evaluate:
   
5. Simplify sin(180o - q ) in terms of sinq or cosq .
   
6. Evaluate sin2(4q ) + cos2(4q ) for all q .
   
7. In a right triangle ABC, angle C is the right angle.
  If side AB = 2 and AC = x, . find an expression for tan B.
 
   
8. Rewrite the trigonometric identity for sin2q and cos2q in terms of the angle q
   
9. Final all solutions of x in the interval 0o £ q < 360o satisfying the equation
  2sin2q + sin2q -1 = 0.
   
10. For what values of q in the interval 0 £ q < 2p is cos4q = 1.
   
11.
What is the period of
   
12. Use the law of cosines given below to find an expression for angle A in
  triangle ABC if AB = 8, AC = 4, and BC = 6.
  Law of cosines: a2 = b2 + c2 – 2bccosA.
 
   
13.
Evaluate:
   
14. Simplify: (cos2q) (tanq) (csc2q )
   
15. Let f(x) = –x2 + 5. Evaluate f(1).
   
16. Find the slope of the line 3x – 5y = 1.
   
17. Write the equation of the line passing through the point (3, –4) having
 
slope
   
18. A rectangle has vertices (7, 7), (10,7), (7, –2) and (10, –2).
  Find the length of the diagonal.
 
   
19.
If f(x) = x2, simplify
   
20. Graph | x | and | x +1| and | x–1|
   
21. If x = e y – 2. Solve for y in terms of x
   
22. The graph of the parabola y = –x2 + 16x + 1 is symmetric with respect to
  what line?
   
23.
If f(x) = 9x2 +1 and Find f(g(x)) and g( f(x)). Simplify if possible.
   
24.
If For which value(s) of x is f(x) = 1?
   
25.
Find the domain and range of
   
26. Find the points of intersection of the graphs y = 2x2 and y = 3-5x.
   
27.
Simplify:
   
28.
Use log rules to simplify .
   
29. The polynomial x(x2 –16)(x2 + 16) has how many real roots?
   
30. Consider y = lnx. What is the range and domain? What is the x intercept?
  Discuss the behavior of the graph as x® ¥ and as x® 0+
   
 

 
Answers for the Review/Tutorial
 
1. Answer
The radian measure for 330o is gotten by multiplying by
which we get by dividing 330 and 180 by 30.
 
2. Answer sinp = 0
and sinp = 0.
     
So the smallest number is sinp = 0.
 
3. Answer 7.8
Since we get
to the nearest tenth.
 
4. Answer
Since is in quadrant III,
the sign of is negative and equal to
So
 
5. Answer sinq .
The trigonometric identity for the expansion of
sin(a – b) = sinacosb – sinbcosa
So that sin(180o – q ) = sin180ocosq - sinq cos180o.
Using sin(180o) = 0 and cos(180o) = –1, we simplify
sin(180o – q ) = 0(cosq ) – sinq (–1) = sinq
 
6. Answer 1
The Pythagorean Identity sin2q + cos2q = 1 applies for any angle.
 
7. Answer
The hypotenuse AB is given as 2, the side AC is given as x
so the third side BC is by using the Pythagorean Theorem.
x2 + (BC)2 = 22, so that (BC)2 = 4 – x2,
Now the
 
8. Answer sin2q = 2sinq cosq   and   cos2q = cos2q – sin2q

Since sin(a + b) = sinacosb + cosasinb and
cos(a + b) = cosacosb – sinasinb,
    by letting a = b = q , we get
sin(q + q ) = sin2q = sinq cosq + cosq sinq = 2sinq cosq .
cos(q + q ) = cos2q = cosq (cosq ) – sinq (sinq ) = cos2q – sin2q .
 
9. Answer q = 30o, 90o, 150o.
Factor: 2sin2q + sinq – 1 = 0 to get
(2sinq – 1)(sinq + 1) = 0
2sinq – 1 = 0
and
 sinq + 1 = 0
sinq = – 1
For there are two solutions in the interval 0 £ q < 2p ,
one in quadrant I and one in quadrant II.
q = 30o and q = 150o
For sinq = – 1, there is one solution, q = 270o.
 
10. Answer
q = 0, p,
For cos4q = 1, 4q = 0 + 2np for integers n or
Letting n range from 0,1,2,3 we get four answers in the required
domain of 0 £ q < 2p .
When n = 4, = 2p > 2p
 
11. Answer
For y = Asin(Bx + C) the period is
   
Thus the period is
 
12. Answer
Using the Law of Cosines with AC = b = 4, BC = a =6, and AB = c = 8,
a2 = b2 + c2 – 2bccosA
becomes 62 = 42 + 82 – 2(4)(8)cosA
36 = 16 + 64 – 64cosA
36 – 80 = – 64cosA
– 44 = – 64cosA
 
13. Answer p
, so that
The arcsinx is the angle q whose sine is x; sinq = x where
 
14. Answer cotq
Using the relations that and
we conclude that
 
15. Answer 4
For f(x) = – x2 + 5,
f(1) = – 12 + 5 = – 1 + 5 = 4. Think of – 12 as – 1(1)2.
Remember the order of operations require that we square 1 first

and then multiply by – 1.
 
16. Answer
Solving for y we get , so the slope is the coefficient of x,
which is
 
17. Answer
The point slope formula for a line passing through the point (xo , yo)
with slope m is y – yo = m(x – xo)
For our problem (xo , yo) = (3 – 4) and
y – yo = m(x – xo)
Thus substituting for xo, yo, m , we get
   
 
18. Answer
The length of the top (and bottom) is 10 – 7 = 3 and the length of the
sides is 7 –(–2) = 7 + 2 = 9. Since the angles of a rectangle are right angles
the diagonal, and the adjacent sides form a right triangle with the diagonal

as the hypotenuse.
The diagonal, d, and the sides satisfy the Pythagorean Theorem.
d2 = 32 + 92 = 9 + 81 = 90
 
19. Answer 2x + a.
If f(x) = x2
 
20. Answer
| x | is defined as x for x ³ 0 which is the right hand line in the first diagram
and as – x for x < 0 which is the left hand line in the first diagram.
In general the graph of f(x – c) shifts the graph of f(x), c units along the
x-axis. Thus | x +1| shifts the graph one unit to the left since x + 1 = x – (–1)
where c = – 1. In a similar way | x– 1| shifts the original graph one unit to the right.
 
21. Answer y = lnx + 2
For x = ey –2, take the natural logarithm of both sides to get
lnx = y – 2,
y = lnx + 2.
 
22. Answer x = 8
The axis of symmetry of a parabola of the form
y = ax2 + bx + c, is Thus
 
23. Answer
f(g(x)) = 9x + 1;
Also Note
because the radical does not distribute over addition.
 
24. Answer x = 1
becomes, after multiplying by x2, 2x –1 = x2
Moving all the terms to one side:
x2 – 2x + 1 = 0
(x – 1)2 = 0
x = 1
 
25. Answer Domain x ³ 4 or x £ – 4, Range y ³ 0
Since the expression under the radical sign must be non-negative we have
x2 –16 ³ 0      x2 ³ 16
| x | ³ 4
x ³ 4 or    x £ – 4
 
26. Answer
x = –3
To find the points of intersection of two curves set the y values equal to each other.
2x2 = 3 – 5x
2x2 + 5x – 3 = 0
(2x –1)(x + 3) = 0
2x –1 = 0 x + 3 = 0
2x = 1 x = –3
 
 
27. Answer – 4
Rewrite
 
28. Answer
The log rules for division and power are
and   ln(ap) = plna we have
 
29. Answer 3 real roots, 0, – 4, 4
x(x2 – 16)(x2 + 16) = 0
x = 0,    x2 – 16 = 0,    x2 + 16 = 0
(x – 4)(x + 4) = 0; x2 = –16 has no real solutions
x = 4, x = – 4
 
30. Answer Domain x > 0, Range – ¥ < y < ¥ , x – intercept is 1.
As x® ¥ , lnx ® ¥ , and as x® 0+, lnx® – ¥ .