Solutions with detailed explanations to compass math test practice questions in sample 4.

If f(x) = 2x  3, g(x) = 2x^{3} + 2 and h(x) = 3x, then h(2) + f(g(1)) =
Solution
We first calculate h(2)
h(2) = 3(2) = 6
We next calculate g(1)
g(1) = 2(1)^{3} + 2 = 2(1) + 2 = 4
We next calculate f(g(1))
f(g(1)) = f(4) = 2(4)  3 = 8  3 = 5
Finally
h(2) + f(g(1)) = 6 + 5 = 11

If 3^{x2} / 3^{4x} = 1/81, then x =
Solution
We first rewrite the two expressions in the given equation to the same base
3^{x2} / 3^{4x} = 3^{x24x}
and
1/81 = 1 / 3^{4} = 3^{4}
We now rewrite the equation as follows
3^{x24x} = 3^{4}
Which leads to the following algebraic equation
x^{2}  4x = 4
x^{2}4x + 4 = 0
factor and solve
(x  2)^{2} = 0
x = 2

What are the solutions of x(x  2) = 5?
Solution
Expand and write in standard form
x^{2}  2x  5 = 0
Use quadratic formulas
a
x = [ (2) + or  sqrt( (2)^{2}  4(1)(5)) ] / 2
= [ 2 + or  sqrt(24) ] / 2
= [ 2 + or  2 sqrt(6) ] / 2
two solutions: 1 + sqrt(6) and 1  sqrt(6)

The formula for the n^{th} term, a_{n}, of an arirhmetic progression is given by a_{n} = a_{1} + (n  1)d, where a_{1} represents the first term of the progession and d represents its common difference. What is the value of the 20^{th} term of the arithmetic progression 4, 7, 10,...?
Solution
The common difference of the given progression is
7  4 = 3 (or 10  7 = 3)
The first term is
a_{1} = 4
The value of the 20^{th} term is given by
a_{20} = a_{1} + (n  1)d = 4 + (20  1)3 = 4 + 3 × 19 = 4 + 57 = 61

In a a 16by12 rectangle, what is the perimeter of the triangle formed by two sides of the rectangle and the diagonal?
Solution
Let x be the length of the diagonal and use Pythagora's theorem to find it.
x^{2} = 16^{2} + 12^{2} = 256 + 144 = 400
x = sqrt(400) = 20
The three sides of the triangle are the length and width of the rectangle and the diagonal.
Perimeter = 16 + 12 + 20 = 48

If sqrt(1) = i, then (2 + 2 i)^{5} =
Solution
We first write the complex number z = (2 + 2i) in exponential form.
z = (2 + 2i) = sqrt((2)^{2} + (2^{2})) e^{[ i arg(2/2) ]} = 2 sqrt(2) e^{[ i arg(2/2) ]}
= 2 sqrt(2) e^{[ 3 Pi/4 i ]}
We now apply De Moivre theorem to find (2 + 2 i)^{5}
(2 + 2 i)^{5} = z ^{5}
= (2 sqrt(2))^{5} e^{[ 5 · 3 Pi/4 i ]}
= 128 sqrt(2) · e^{[ 15 Pi/4 i ]}
= 128 *sqrt(2)* [ sin(15 Pi/4) + i cos(15 Pi/4) ]
= 128 sqrt(2) [1/sqrt(2)  i * 1/sqrt(2) ] = 128  128 i

For all values of x > 0, log[( x^{2} cube root(7) ] =
Solution
We first the log formula log (A*B) = log A + log B to write that
log[( x^{2} cube root(7) ] = log( x^{2}) + log(cube root(7))
We now use the log formula log (A^{n}) = n log A to simplify the above
= 2 log x + log (7^{(1/3)})
= 2 log x + (1/3) log 7

Which of these interval represents all the real values that are the range of y = 1 / (4  x^{2})
Solution
y = 1 / (4  x^{2}) is a even function since
y(x) = 1 / (4  (x)^{2}) = 1 / (4  x^{2})
Because it is an even function and therefore its graph is symmetric with respect to the y axis, we shall study its graph for x ≥ 0.
Since the denominator of this rational function is zero at x = + or 2, it has 2 vertical asymptotes: x = 2 and x =  2.
It also has a horizontal asymptote given by y = 0 because the degree of the denominator is greater than the degree of the numerator.
For 0 ≤ x < 2, 4  x^{2} is positive and therefore 1 / (4  x^{2}) is also positive.
For x > 2 , 4  x^{2} is negative and therefore 1 / (4  x^{2}) is also negative.
Also for x = 0, y = 1/4
Putting all the above information together to graph the given function, we end up with the graph below.
.
The range of the given function is given by the interval.
(infinity , 0) U [1/4 , +infinity)

Which of these intervals represents all values of x that makes sqrt(x^{2} + 2x) a real number?
Solution
sqrt(x^{2} + 2x) is real if .
x^{2} + 2x ≥ 0
or
x(x + 2) ≥ 0
x(x + 2) changes sign at the values of x that make x(x + 2) = 0 which are x = 0 and x =  2. Hence 3 intervals to study the sign of x(x + 2)
interval 1: ( infinity , 2] , x(x + 2) ≥ 0
interval 2: [2 , 0] , x(x + 2) ≤ 0
interval 3: [0 , +infinity) , x(x + 2) ≥ 0
sqrt(x^{2} + 2x) is real for x in the interval
(infinity , 2] U [0 , +infinity)

log 32 + log 16 =
Solution
Note that
32 = 2^{5} and 16 = 2^{4}
Hence
log 32 + log 16 = log 2^{5} + log 2^{4}
= 5 log 2 + 4 log 2 = 9 log 2

If 9^{(x + 1)} = 3 * 9^{y}, then
Solution
Let us rewrite the two sides of the equation to the same base
9^{(x + 1)} = (3^{2})^{(x + 1)} = 3^{2(x + 1)}
3 * 9^{y} = 3^{1} * (3^{2})^{y} = 3^{1 + 2y}
We now rewrite the given equation as follows
3^{2(x + 1)} = 3^{1 + 2y}
Which leads to
2(x + 1) = 1 + 2y
Solve for x
2x + 2 = 1 + 2y
2x = 2y  1

If A is a matrix given by
, then
Solution
The determinant of a 2 by 2 matrix of the form _{} is given by
AD  BC
Apply the above to the given matrix to obtain the determinant as follows
a(a)  (ad)(e) = a^{2}  aed = a(a + ed)

If cos(80^{o}) = a and cos(60^{o})cos(20^{o}) = b, then sin(60^{o})sin(20^{o}) =
Solution
Write cos(80^{o}) as follows
cos(80^{o}) = cos(60^{o} + 20^{o})
Expand
= cos(60^{o})cos(20^{o})  sin(60^{o}) sin(20^{o})
Substitute cos(80^{o}) by a and cos(60^{o})cos(20^{o}) by b
a = b  sin(60^{o}) sin(20^{o})
Hence
sin(60^{o}) sin(20^{o}) = b  a

Which pair of functions have the same graph?
A) sin(x) and cos(x + 3 Pi/2)
B) sin(x) and cos(x + Pi/2)
C) sin(x) and cos(x + Pi)
D) sin(x) and cos(x  3 Pi/2)
E) sin(x) and cos(x + 5 Pi/2)
Solution
Let us expand cos(x + 3 Pi/2) and simplify
cos(x + 3 Pi/2) = cos(x) cos(3 Pi/2)  sin(x) sin(3 Pi/2) = cos(x) 0  sin(x)(1) = sin(x)
Since cos(x + 3 Pi/2) simplify to sin(x), the two functions have the same graph.
NOTE: As an exercise, expand the remaining functions: cos(x + Pi/2)
, cos(x + Pi), cos(x  3 Pi/2) and cos(x + 5 Pi/2) and show that none is equal to sin(x).

Which of these functions have the highest period?
A) cos(x + 3 Pi/2)
B) sin(2x  Pi)
C) cos(0.2x)
D) 10 sin(x)
E) cos(200x)
Solution
The period of a function of the form y = a sin(bx + c) or y = a cos(bx + c) is given by
2 Pi / b
Let us now find the periods of the given functions
A) cos(x + 3 Pi/2) : Period = 2 Pi / 1 = 2 Pi
B) sin(2x  Pi) : Period = 2 Pi / 2 = Pi
C) cos(0.2x) : Period = 2 Pi / 0.2 = 10 Pi
D) 10 sin(x) : Period = 2 Pi / 1 = 2 Pi
E) cos(200x) : Period = 2 Pi / 200 = Pi / 100 = 0.01 Pi
The function with the highest period is
cos(0.2x)

What is the measure of x, 0 < x < 5 Pi / 2, if  sin(x) + 1 = 2
Solution
If  sin(x) + 1 = 2 , then
sin(x) + 1 = 2 or sin(x) + 1 = 2
sin(x) =  1 or sin(x) = 3
The equation sin(x) = 3 has no real solutions. However the solution for the equation sin(x) =  1 such that 0 < x < 5 Pi / 2 is
x = 3 Pi/2

If Pi < x < 2 Pi and cos(x) =  1 / 2, then x =
Solution
Start with
cos(Pi / 3) = 1/2
For Pi < x < 2 Pi and cos(x) =  1/2, x is in quadrant III. Hence
x = Pi + Pi/3 = 4 Pi/3

[ cos(t) / cot(t) ] csc^{2}(t) =
Solution
Use the identities csc(t) = 1 / sin(t) and tan(t) = 1 / cot(t) to write
[ cos(t) / cot(t) ] csc^{2}(t) = cos(t) tan(t) / sin^{2}(t)
= cos(t) [ sin(t) / cos(t)] / sin^{2}(t)
simplify
= 1 / sin(t)= csc(t)

If f(u) = 2 cos(u) + 5 and g(v) = 0.5 sin(2v), what is f(g(Pi / 2))
Solution
We first calculate g(Pi/2)
g(Pi/2) = 0.5 sin(2 Pi/2) = 0
f(g(Pi/2)) is calculated as follows
f(g(Pi / 2)) = f(0) = 2 cos(0) + 5 = 2 * 1 + 5 = 7

If Pi < x < 2 Pi and tan(x) = 1/4, then sin(x) =
Solution
Use the definition of tan(x) = b / a where point (a,b) is on the terminal side of angle x to write
tan(x) = 1/4 = b / a
Since x is in quadrant III, we can write
b =  1 and a =  4
r distance from (0,0) to (a,b) is given by
r = sqrt(1^{2} + 4^{2}) = sqrt(17)
We now use sin(x) = b / r
sin(x) = 1 / sqrt(17)
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