__Questions 1:__

If Log_{x} (1 / 8) = - 3 / 2, then x is equal to

__Solution__

Rewrite the logarithmic equations using exponential equation

x^{- 3 / 2} = 1 / 8

The above equation may be rewritten as

x^{ - 3 / 2} = 1 / 2^{3} = 2^{-3}

(√x)^{-3} = 2^{-3}

The above equation gives

√x = 2 or x = 4

__Question 2:__

20 % of 2 is equal to

__Solution__

20 % of 2 = (20 / 100) * 2 = 0.4

__Questions 3:__

If Log_{ 4} (x) = 12, then log_{ 2} (x / 4) is equal to

__Solution__

If Log_{ 4} (x) = 12, then

x = 4^{12}

x / 4 = 4^{11}

Hence

log_{ 2} (x / 4) = log_{ 2} (4^{11})

= log_{ 2} ((2^{2})^{11})

= log_{ 2} (2^{22})

= 22

__Questions 4:__

The population of a country increased by an average of 2% per year from 2000 to 2003. If the population of this country was 2 000 000 on December 31, 2003, then the population of this country on January 1, 2000, to the nearest thousand would have been

__Solution__

A population of size P increasing at the rate of 2% may be modelled as follows

P = P_{0} e^{0.02 t} , where t is the number of years after t = 0 and P_{0} is the population at t = 0.

Let t = 0 corresponds to January 1, 2000 and therefore t = 4 corresponds to December 31. But P is 2,000,000 when t = 4. Hence

2,000,000 = P_{0} e^{0.02*4}

Solve the above for P_{0}

P_{0} = 2,000,000 / e^{0.02*4} = 1 846 000 (rounded to the nearest thousand)

P_{0} is the population at t = 0 or on January 1, 2000.

__Questions 5:__

f is a quadratic function whose graph is a parabola opening upward and has a vertex on the x-axis. The graph of the new function g defined by g(x) = 2 - f(x - 5) has a range defined by the interval

__Solution__

If the graph of f is a parabola with vertex on x axis and opening up, then the range of f is given by the interval

[0 , + infinity)

The graph of f(x - 5) is that of f(x) shifted 5 units to the right and therefore no change to the range. However since the graph of - f(x - 5) is that of f(x - 5) reflected on the x axis then the range of - f(x - 5) is given by

(- infinity , 0]

The graph g(x) = 2 - f(x - 5) is that of - f(x - 5) shifted up by 2 units; hence the range of g(x) is given by

(- infinity , 2]

__Questions 6:__

f is a function such that f(x) < 0. The graph of the new function g defined by g(x) = | f(x) | is a reflection of the graph of f

A. on the y axis

B. on the x axis

C. on the line y = x

D. on the line y = - x

__Solution__

Note that since f(x) < 0 then

g(x) = | f(x) | = - f(x)

The graph of g(x) = - f(x) is therefore the reflection of the graph of f on the x axis.

__Questions 7:__

If the graph of y = f(x) is transformed into the graph of 2y - 6 = - 4 f(x - 3), point (a , b) on the graph of y = f(x) becomes point (A , B) on the graph of 2y - 6 = - 4 f(x - 3) where A and B are given by

A. A = a - 3, B = b

B. A = a - 3, B = b

C. A = a + 3, B = -2 b

D. A = a + 3, B = -2 b +3

__Solution__

We first solve 2y - 6 = - 4 f(x - 3) for y.

y = - 2 f(x - 3) + 3

The graph of y = - 2 f(x - 3) + 3 is that of y = f(x) shifted 3 units to the right, stretched vertically by a factor of 2, reflected on the x axis and shifted up by 3 units. A point of y = f(x) will undergo the same transforamtions. Hence

Point (a , b) on the graph of y = f(x)

Becomes (a + 3 , b) on the graph of y = f(x - 3) : shifted 3 units to the right

Becomes ( a + 3 , 2 b) on the graph of y = 2 f(x - 3) : stretched vertically by 2

Becomes ( a + 3 , - 2 b) on the graph of y = - 2 f(x - 3): reflected on x axis

Becomes ( a + 3 , - 2 b + 3) on the graph of y = - 2 f(x - 3) + 3 : shifted up 3 units

__Questions 8:__

When a parabola represented by the equation y - 2x^{ 2} = 8 x + 5 is translated 3 units to the left and 2 units up, the new parabola has its vertex at

__Solution__

First rewrite y - 2x^{ 2} = 8 x + 5 as

y = 2x^{ 2} + 8 x + 5

Complete square and determine vertex.

y = 2(x^{ 2} + 4x + 4) - 8 + 5

= 2(x + 2)^{ 2} - 3

vertex at (- 2 , - 3)

If parabola is translated 3 units to the left and 2 units up its vertex is also translated 3 units to the right and 2 units up .

vertex after translations is at: (-2 - 3 , - 3 + 2) = (-5 , -1)

__Questions 9:__

The graphs of the two linear equations ax + by = c and bx - ay = c, where a, b and c are all not equal to zero,

A. are parallel

B. intersect at one point but not perpendicular

C. intersect at two points

D. perpendicular

__Solution__

let us find the slopes of the two lines

a x + b y = c , slope m1 = - a / b

b x - a y = c , slope m2 = b / a

m1*m2 = (- a / b)(b / a) = - 1

The two lines are perpendicular

__Questions 10:__

The graphs of the two equations y = a x^{ 2} + b x + c and y = A x^{ 2} + B x + C, such that a and A have different signs and that the quantities b^{ 2} - 4 a c and B^{ 2} - 4 A C are both negative,

A. intersect at two points

B. intersect at one point

C. do not intersect

D. none of the above

__Solution__

Since a and A have different signs the graphs of the two equations are parabolas opening in diffent directions: If one opens up the other opens down. Also since b^{ 2} - 4 a c and B^{ 2} - 4 A C are both negative, none of the parabola cuts the x axis. This means that each one of these parabolas is either above the x axis or below the x axis and therefore do not intersect.

__Questions 11:__

For x greater than or equal to zero and less than or equal to 2 pi, sin x and cos x are both decreasing on the intervals

A. (0 , pi/2)

B. (pi/2 , pi)

C. (pi , 3 pi / 2)

D. (3 pi / 2 , 2 pi)

__Solution__

sin(0) = 0 and cos(0) = 1, and from x = 0 to x = pi/2, sin(x) increases from 0 to 1 and cos(x) decreases from 1 to 0. From x = pi/2 to x = pi, sin(x) decreases from 1 to 0 and cos(x) decreases from 0 to -1. Hence both sin(x) and cos(x) decreases on the interval (pi/2 , pi)

__Questions 12:__

The three solutions of the equation f(x) = 0 are -2, 0, and 3. Therefore, the three solutions of the equation f(x - 2) = 0 are

__Solution__

If f(x) = 0 at x = -2, 0 and 3 then f(x - 2) = 0 for

x - 2 = -2 , x - 2 = 0 and x - 2 = 3

Solve the above equations to find

x = 0 , x = 2 and x = 5

__Questions 13:__

The three solutions of the equation f(x) = 0 are - 4, 8, and 11. Therefore, the three solutions of the equation f(2 x) = 0 are

__Solution__

If f(x) = 0 at x = - 4, 8 and 11 then f(2x) = 0 for

2x = -4 , 2x = 8 and 2x = 11

Solve the above equations to find

x = -2 , x = 4 and x = 11/2

__Questions 14:__

A school committee consists of 2 teachers and 4 students. The number of different committees that can be formed from 5 teachers and 10 students is

__Solution__

There are C(5,2) ways to select 2 teachers from 5 and C(10,4) ways to select 4 students from 10 where C(n,r) is the combinations of n items taken r at the time. Using the multiplication counting principle,the number of different committees that can be formed is given by

C(5,2)*C(10,4) = 2100

__Questions 15:__

Five different books (A, B, C, D and E) are to be arranged on a shelf. Books C and D are to be arranged first and second starting from the right of the shelf. The number of different orders in which books A, B and E may be arranged is

__Solution__

Since books C and D are arranged first and second, only books A, B and E will change order. Therefore it an arrangement problem involving 3 items and the number of different order is given by

3!

__Questions 16:__

The mean of a data set is equal to 10 and its standard deviation is equal to 1. If we add 5 to each data value, then the mean and standard deviation become

__Solution__

Since 5 is added to all data values, the mean will also increase by 5 and becomes 15. But the standard deviation which measure the "distance" between the mean and the data values does not change.

__Questions 17:__

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

__Solution__

Let m be the mean and s be the standard deviation and find the z score.

z = (x - m) /s = (0.8 s + m - m) / s = 0.8

The percentage of student who scored above Jane is (from table of normal distribution).

1 - 0.7881 = 0.2119 = 21.19%

The number of student who scored above Jane is (from table of normal distribution).

21.19% 0f 500 = 106

__Questions 18:__

If f(x) is an odd function, then | f(x) | is

A. an odd function

B. an even function

C. neither odd nor even

D. even and odd

__Solution__

Since f(x) is odd, we have

f(-x) = - f(x)

Let g(x) = |f(x)|.

g(- x) = |f(- x)| = |- f(x)) = |f(x)| = g(x)

|f(x)| is even.

__Questions 19:__

The period of | sin (3x) | is

__Solution__

The period of sin (3x) is given by

2 pi / 3

In order to graph |sin(3x)| given the graph of sin(3x), we need to reflect all neagtive parts of the graph of sin(3x) which halves the period. Hence the peiod of |sin(3x)| is given by

(2 pi / 3) / 2 = pi / 3

__Questions 20:__

When a metallic ball bearing is placed inside a cylindrical container, of radius 2 cm, the height of the water, inside the container, increases by 0.6 cm. The radius, to the nearest tenth of a centimeter, of the ball bearing is

__Solution__

When the ball was put in water, the volume of water increased by.

pi*2^{2}*0.6 = 2.4 pi cubic cm

The above volume is equal to the volume of the ball with radius r to find. Hence

(4/3)*pi*r^{3} = 2.4 pi

Solve the above for r

r = 1.2 cm

__Questions 21:__

The period of 2 sin x cos x is

__Solution__

According to trigonometric identities, 2 sin x cos x = sin(2x) and the period is given by.

2 pi / 2 = pi

__Questions 22:__

The probability that an electronic device produced by a company does not function properly is equal to 0.1. If 10 devices are bought, then the probability, to the nearest thousandth, that 7 devices function properly is

__Solution__

Let q being the probability a device does not function properly and p = 1 - q = 0.9 the probability that a device functions properly. Since there are only two possible results, it is a binomial distribution. The P probability that 7 out of 10 devices function properly is given by

C(10,7)*0.9^{ 7}*0.1^{ 3} , where C(10,7) is the number of ways that 7 items are selected from 10 and is given by

P = C(10,7) = 10! / (7!3!)

A calculation of P gives

P = 0.057

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