MATH 1090 Algebra and Trigonometry
Final Examination - Solutions

2002 Winter

Note

Perform the indicated operation and simplify:
2. The LCD (lowest common denominator) is x (x - 1)³
3. OR
  One may use the difference of squares formula first:

Factor:
2. Factor by grouping; then use :

Solve the given equation or inequality.

LCD = (x + 2)(x + 3)²
Multiplication of the equation by its LCD is valid provided LCD ¹ 0;
that is, for all x except -2 and -3.
(x + 3)² + (x + 2)(x + 3) = 6(x + 2)
Þ (x² + 6x + 9) + (x² + 5x + 6) = 6x + 12
Þ 2x² + 5x + 3 = 0
Þ (2x + 3) (x + 1) = 0
Þ x = -3/2 or -1
A quick check confirms that each of these values of x does indeed satisfy the original equation.
The solution set is therefore

Squaring both sides (which may introduce extraneous solutions):
x² - 2x + 1 = x + 5
Þ x² - 3x - 4 = 0
Þ (x - 4)(x + 1) = 0
Þ x = 4 or -1

Checking in the original equation:
x = 4 Þ LHS = x = 4

Þ x = 4 is a solution.

x = -1 Þ LHS = x = -1

Þ x = -1 is not a solution.
[x = -1 is the solution to the different equation .]
The solution set is therefore

| 4x - 3 | > 9
Þ 4x - 3 > 9 OR 4x - 3 < -9
Þ 4x > 12 or 4x < -6
Þ x > 3 or x < -3/2
Therefore the solution set is

x (2x + 3) < 2

Þ 2x² + 3x - 2 < 0
Þ (2x - 1)(x + 2) < 0
Critical numbers: 1/2 and - 2

Interval	2x - 1	x + 2	(2x - 1)(x + 2)	ineq. true?
(-¥, -2)	-	-	+	no
(-2, 1/2)	-	+	-	yes
(1/2, ¥)	+	+	+	no

The original inequality is therefore true only on the middle interval.
The solution set is, therefore,
[-2, 1/2]

[Note: as the original inequality is a quadratic inequality, it is sufficient to test any one of the three test intervals. The simplest test is at x = 0 , at which the inequality is obviously true ( 0 × (2(0) + 3) < 2 ). The conclusion above then follows, more efficiently.]

Find an equation of the line passing through the point (2, 3) and perpendicular to the line 2x + 3y + 1 = 0 .
The given line is 2x + 3y + 1 = 0

, where m₁ is the slope of the given line.
The required line is perpendicular to the given line
Þ the slope of the required line, m₂ , is given by

The point (2, 3) is known to lie on the required line.
Therefore, in point-slope form, an equation of the required line is
[This can be re-written as

Sketch the graph of each of the following equations.
Label all intercepts and also label the vertex, if appropriate.
1. y = | 2x - 3 | - 4
  
  This is the graph of y = | x | after
  • a horizontal shift of 3/2 to the right, then
  • a vertical stretch × 2, then
  • a vertical shift of 4 down.
  The graph therefore opens upward, with its vertex at (3/2, -4).
  Axis intercepts:
  x = 0 Þ y = | 0 - 3 | - 4 = -1
  y = 0 Þ | 2x - 3 | - 4 = 0 Þ | 2x - 3 | = 4
  Þ 2x - 3 = ±4 Þ 2x = 3 ±4 = -1 or +7
  Þ x = -1/2 or +7/2
  The sketch then follows:
2. y = -x² + 8x - 12
  y = -x² + 8x - 12
  Compare this to the standard form y = ax² + bx + c
  Þ a = -1 (a < 0 Þ parabola opens downward);
  b = +8 , c = -12
  
  Þ k = -h² + 8h - 12 = -16 + 32 - 12 = 4
  The vertex is therefore at (h, k) = (4, 4)
  [OR complete the square: y = -(x - 4)² + 4 .]
  Axis intercepts:
  x = 0 Þ y = - 12
  y = 0 Þ -x² + 8x - 12 = 0
  Þ x² - 8x + 12 = 0
  Þ (x - 2) (x - 6) = 0
  Þ x = 2 or 6
  The sketch then follows:

For the functions and ,
find and the domain of .

Therefore
The domain of is all x that do not cause the denominator to be zero in either or g(x) .
The domain is therefore
$D_(fog) = { x | x not= 0, x not= 2 }$

If , find f ^-1 (x) and sketch the graphs of f and f ^-1 on the same set of axes.

$x = sqrt{1-2y} ==> x^2 = 1 - 2y$

Therefore
The domain of f ^-1 is the same as the range of f.
Therefore
$D_(f^(-1)) = { x | x >= 0 }$ It is easier to sketch the graph of f ^-1 first.
It is the right half only of the downward opening parabola that has its vertex at and its positive x intercept at (1, 0). The graph of f is the reflection in the line y = x of the graph of f ^-1. The sketch then follows:

1. Use the remainder theorem to find the remainder when f (x) = x⁷ + 3x⁶ - 5x³ - 5x² + 7 is divided by .
  The remainder R upon dividing f (x) by x - c is R = f (c) (remainder theorem).
  $f(x) = x^7 + 3x^6 - 5x^3 - 5x^2 + 7 ==> f(sqrt{2}) = 2^(7/2) + 3(2)^(6/2) - 5(2)^(3/2) - 5(2)^(2/2) + 7$
  $= 2^3 sqrt{2} + 3(2^3) - 5(2)sqrt{2} - 5(2) + 7$
  
  Therefore
2. Find c so that x + 2 is a factor of g (x) = cx⁵ + 2x³ - 8cx - 32 .
  In order for x + 2 to be a factor of g (x) , the remainder R = g (-2) must be zero.
  g (-2) = -32 c - 16 + 16 c - 32 = 0
  Þ -48 = 16 c
  Therefore
  
  A less efficient alternative is to apply synthetic division, then to require that the remainder be zero:
  
  R = -16c - 48 = 0 Þ c = -3

For the polynomial P(x) = 9x⁴ - 6x³ + 37x² - 24x + 4:

List all possible rational zeroes.
The complete list consists of all fractions (positive or negative) that can be formed by selecting a numerator from the integer factors of the constant coefficient (4) and a denominator from the integer factors of the leading coefficient (9). The full list is therefore
However, by Descartes’ rule of sign, the pattern of alternating signs in the coefficients allows us to conclude that none of the zeroes of P(x) can be negative. The list of possible zeroes may therefore be reduced to

Factor P(x) completely into linear factors.

An obvious first attempt fails:

1 \|	9	-6	37	-24	4
		9	3	40	16

	9	3	40	16	20

There are no sign changes on the last row Þ no zeroes are greater than 1.
Five of the original 18 possibilities remain. The only one that produces a solution is:

1/3 \|	9	-6	37	-24	4
		3	-1	12	-4

	9	-3	36	-12	0

The cubic factor may be factored further by grouping:
P(x)= (3x-1)(x^2 (3x-1) + 4(3x-1))
= (3x-1)^2 (x^2 + 4)

Therefore

If one does not spot the factoring by grouping, then one may attempt a second factoring by synthetic division:

1/3 | 9 -6 37 -24 4

3 -1 12 -4

1/3 | 9 -3 36 -12 0

3 0 12

9 0 36 0

P(x) = (x-1/3)^2 (9x^2 + 36)
= 9(x-1/3)^2 (x^2 + 4)
The same final form then follows.

Solve for x:
1. Note that 8 and 16 are both integer powers of 2.
  
  Þ 6x - 3 = 4x + 5 Þ 2x = 8
  Therefore
  
  OR
  Taking logarithms (to any one valid base) on both sides of the equation:
  
  Þ (2x - 1) log 8 = log 2 + (x + 1) log 16
  But log 8 = log (2³) = 3 log 2
  and log 16 = log (2⁴) = 4 log 2
  Þ (2x - 1) 3 log 2 = log 2 + (x + 1) 4 log 2
  Þ 3 (2x - 1) = 1 + 4(x + 1)
  Þ 6x - 3 = 4x + 5 Þ 2x = 8
  Therefore
2. $1 + (1/2)log_2 {x-1} = log_2 {x-4}$
  $1 = log_2 {x-4} - (1/2)log_2 {x-1} = log_2 {(x-4)/sqrt(x-1)}$
  $2^1 = (x-4)/sqrt{x-1} ==> 2 sqrt{x-1} = x-4$
  Þ 4(x - 1) = x² - 8x + 16
  Þ x² - 12x + 20 = 0
  Þ (x - 2) (x - 10) = 0
  Þ x = 2 or x = 10
  However, squaring both sides may have introduced extraneous solutions. Checking:
  x = 2:
  LHS = 1 + ½ log₂1 = 1
  RHS = log₂ (2 - 4) = not real
  
  x = 10:
  LHS = 1 + ½ log₂9 = 1 + log₂3
  = log₂2 + log₂3 = log₂6
  RHS = log₂ (10 - 4) = log₂6 = LHS
  Therefore

Find all values of q in the interval [0, 2p) that satisfy the equation sin 2q + sin q = 0 .
sin 2q + sin q = 0 Þ 2 sin q cos q + sin q = 0
Þ sin q (2 cos q + 1) = 0
Þ sin q = 0 or 2 cos q + 1 = 0
Þ sin q = 0 or cos q = -½
But cos 60° = +½ and cos q < 0 Þ q is in quadrant II or III.
Also sin (np) = 0 for all integers n.
Therefore the solution set for q (in radians) is:

Given that and that , find the exact value of:
1. sin q
  
  It follows immediately that
2. tan (q + 2p)
  tan (q + 2p) = tan q (for any q)
  From the quadrant diagram,
3. Therefore
4. cos 2q
  Three different equally valid methods are presented here:
  
  OR
  
  OR
  
  Therefore
5. One half of any fourth quadrant angle is an angle in the [second half of the] second quadrant, where the cosine ratio is negative.
  Therefore we need to take the negative square root in the half-angle formula:
  
  Therefore