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Solutions with Explanations - Sample 4

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

If f(x) = 2x - 3, g(x) = -2x³ + 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^x²-4x

and

1/81 = 1 / 3⁴ = 3^-4

We now rewrite the equation as follows

3^x²-4x = 3^-4

Which leads to the following algebraic equation

x² - 4x = -4

x²-4x + 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

x² - 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 = a₁ + (n - 1)d, where a₁ 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

a₁ = 4

The value of the 20^th term is given by

a₂₀ = a₁ + (n - 1)d = 4 + (20 - 1)3 = 4 + 3 × 19 = 4 + 57 = 61
In a a 16-by-12 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.

x² = 16² + 12² = 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 + (2²)) 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[( x² cube root(7) ] =

Solution

We first the log formula log (A*B) = log A + log B to write that

log[( x² cube root(7) ] = log( x²) + log(cube root(7))

We now use the log formula log (Aⁿ) = 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 - x²)

Solution

y = 1 / (4 - x²) is a even function since

y(-x) = 1 / (4 - (-x)²) = 1 / (4 - x²)

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 - x² is positive and therefore 1 / (4 - x²) is also positive.

For x > 2 , 4 - x² is negative and therefore 1 / (4 - x²) 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(x² + 2x) a real number?

Solution

sqrt(x² + 2x) is real if .

x² + 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(x² + 2x) is real for x in the interval

(-infinity , -2] U [0 , +infinity)
log 32 + log 16 =

Solution

Note that

32 = 2⁵ and 16 = 2⁴

Hence

log 32 + log 16 = log 2⁵ + log 2⁴

= 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)} = (3²)^{(x + 1)} = 3^{2(x + 1)}

3 * 9^y = 3¹ * (3²)^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) = -a² - 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) ] csc²(t) =

Solution

Use the identities csc(t) = 1 / sin(t) and tan(t) = 1 / cot(t) to write

[ cos(t) / cot(t) ] csc²(t) = cos(t) tan(t) / sin²(t) = cos(t) [ sin(t) / cos(t)] / sin²(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(1² + 4²) = sqrt(17)

We now use sin(x) = b / r

sin(x) = -1 / sqrt(17)

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Free Compass Math Test Practice Questions Solutions with Explanations - Sample 4

Free Compass Math Test Practice Questions
Solutions with Explanations - Sample 4