__Questions 1:__

If f(x) = Log(x), then f^{ -1}(x) =

__Solution__

y = Log(x) if and only if

x = 10^{y}

Interchange x and y

y = 10^{x}

Hence

f^{ -1}(x) = 10^{x}

__Questions 2:__

Log_{4} 65 =

__Solution__

Use the change of base formula to write

Log_{4} 65 = ln(65) / ln(4)

Then use a calculator

= 3.01 (rounded to 2 decimal places)

__Questions 3:__

If 2^{ 3x - 1} = 16, then x =

__Solution__

Rewrite equation using 16 = 2^{4}

2^{ 3x - 1} = 2^{4}

The two bases of the exponential expressions in the above equation are equal to 2. Hence their exponents has also to be equal

3x - 1 = 4

Solve for x

x = 5/3

__Questions 4:__

If Log_{x} 9 = 2, then x =

__Solution__

Log_{x} 9 = 2 if and only if

x^{ 2} = 9

x^{ 2} = 3^{2}

For the above exponential expressions to be equal, their bases must be equal. Hence

x = 3

__Questions 5:__

If x^{ 2} + kx - 6 = (x - 2)(x + 3), then k =

__Solution__

Expand the polynomial on the right side

x^{ 2} + kx - 6 = x^{ 2} + x - 6

For two polynomials to be equal, all their corresponding coefficients must be equal. Hence

k = 1

__Questions 6:__

The vertex of the graph of y = 2x^{ 2} + 8x - 3 is given by

__Solution__

Write y in vertex form by completing the square

y = 2x^{ 2} + 8x - 3 = 2(x^{ 2} + 4x + 4 - 4) - 3

= 2(x^{ 2} + 4x + 4) - 8 - 3

= 2(x + 2)^{2} - 11 which is of the form a(x - h)^{2} + k

vertex at (-2 , -11)

__Questions 7:__

The quadratic equation whose roots are at x = 3 and x = 5 is given by

A. (x - 3)(x - 5) = 1

B. (x + 3)(x + 5) - 9 = (x + 3)(x + 5) - 25

C. (x + 3)(x + 5) = 0

D. x^{ 2} - 8x = -15

__Solution__

Checking the given roots by substituting in equations A., B., C. and D. shows that D. is the equation with roots 3 and 5.

__Questions 8:__

The roots of the equation x / (x + 2) + 3 / (x - 4) = (4x + 2) / (x^{ 2} - 2x - 8) are

__Solution__

We first note that

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

We now multiply all terms of the given equation by (x + 2)(x - 4) and simplify to get

x (x - 4) + 3 (x + 2)= (4x + 2)

Expand, group and solve

x^{ 2} - x + 6 = 4x + 2

x^{ 2} - 5x + 4 = 0

(x - 1)(x - 4) = 0

Roots: x = 1 and x = 4

Check answer: only x = 1 is a solution since the given equation is not defined at x = 4.

__Questions 9:__

If the domain of function f given by f(x) = -x^{ 2} + 6x is given by the interval [0 , 6], then the range of f is given by the interval

__Solution__

We first note that the zeros of f are x = 0 and x = 6. The graph of f cuts the x axis at x = 0 and x = 6. Also since the leading coefficient is negative, the graph of f which is a parabola opens downward and the maximum value of f(x) is within the interval [0,6]. The maximum of f(x)is at

x = - b / (2a) = - 6 / -2 = 3

the maximum value of f(x) is equal to

f(3) = -(3)^{ 2} + 6(3) = 9

The range of f is given by interval

[0 , 9]

__Questions 10:__

The x intercepts of the graph of y = -x^{ 2} + 3x + 18 are given by

__Solution__

The x intercepts are found by solving the equation

- x^{ 2} + 3x + 18 = 0

-(x^{ 2} - 3x - 18) = 0

-(x - 6)(x + 3) = 0

solutions: x = 6 and x = -3

x intercepts: (6 , 0) and (-3 , 0)

__Questions 11:__

The domain of f(x) = 2 Log |x - 2| is given by the interval

__Solution__

The domain of f is the set of x values for which f is defined. f is defined for

|x - 2| > 0

|x - 2| is positive for all x except x = 2. Hence the domain is given by the interval

(-infinity , 2) U (2 , + infinity)

__Questions 12:__

If 1.56^{x} = 2, then x =

__Solution__

Take the ln of both sides

ln(1.56^{x}) = ln(2)

x ln(1.56) = ln(2)

Take the ln of both sides

x = ln(2) / ln(1.56)

__Questions 13:__

The reference angle to angle a = -1280° is equal to

__Solution__

How many -360° are there in -1280°?

-1280/-360 = 3.5 (approximately)

Let us now find a coterminal angle to -1280° by adding 4×360°

-1280 + 4×360 = 160°

Angle of measure 160° has terminal side in quadrant II. Hence the reference angle is given by

180 - 160 = 20°

__Questions 14:__

If x is an angle in standard position with point A(-3 , 4) on the terminal side, then sec x =

__Solution__

Let r be the distance from the (0,0) to (-3,4).

r = √((-3)^{2} + 4^{2}) = 5

sec x = 1/cos x = 1 / (-3/5) = -5/3

__Questions 15:__

If x is an angle in standard position and it terminal side is in quadrant IV and is given by y = - x, then sin x =

__Solution__

Select a point in quadrant IV and on the line y = - x.

(3,-3)

let r be the distance from (0,0) to point (3,-3).

r = √(3^{2} + (-3)^{2}) = 3 √(2)

sin x = -3 / 3 √(2) = - 1 / √(2)

__Questions 16:__

Which statement is NOT true?

A. cos(-x) = cos x

B. tan(-x) = tan x

C. the amplitude of y = -2 cos (t) is equal to 2

D. the range of y = -5 sin (t) is given by [-5 , 5]

__Solution__

B is not true.

tan (- x) = - tan x , odd function

__Questions 17:__

Which statement is true?

A. sec(-x) = - sec x

B. the range of y = tan x is given by (0 , + infinity)

C. the period of y = -2 cos (3 π t) is equal to 2π / 3

D. the period of y = -5 tan (0.5 π t) is equal to 2

__Solution__

D is true.

If y = a tan(bx) then period = π/ |b| = π / (0.5 π) = 2

__Questions 18:__

The radian measure of an angle of 25° is equal to

__Solution__

25 × π / 180 = 5 π / 36

__Questions 19:__

The period of the function f(x) = sin(x π/6 + π/4) is equal to

__Solution__

2π / (π/6) = 12

__Questions 20:__

Which of the following is an identity?

A. cos(2x) = 2 cos x

B. cos(x + y) = cos x + cos(y)

C. sin(x - y) = sin x - sin(y)

D. sin(2x) = 2 sin x cos x

__Solution__

D. For all x real, sin(2x) = 2 sin x cos x

__Questions 21:__

Which of the following is NOT an identity?

A. cot(x + y) = [ 1 + cot x cot(y) ] / [ cot x + cot(y) ]

B. tan(x + y) = [ tan x + tan(y) ] / [ 1 - tan x tan(y) ]

C. sin(x - y) = sin x cos(y) - cos x sin(y)

D. cos(2x) = 2 cos^{ 2}(x) - 1

__Solution__

A. is not an identity.

__Questions 22:__

If x is an angle such that tan x = 5/12 and π < x < 3 π/2, then sec x =

__Solution__

tan x = b/a, where a and b are the coordinates of a point on the terminal side. Since x is in quadrant III, a and b are negative. Hence

tan x = a / b = 5/12 = (-5) / (-12) , a = -12 and b = -5

If r is the distance from (0,0) to (-5,-12), then

r = √((-12)^{2} +(-5)^{2}) = 13

sec x = 1 / cos x = 1 / (-12/13) = -13/12

__Questions 23:__

The real solutions to the equation cos^{ 2}(x) - 1.5 cos x = 1 are given by the solutions to the equation

A. cos x = 1

B. cos x = 2

C. cos x = 1/2

D. cos x = - 1/2

__Solution__

Let's first rewrite the given equation as follows

cos^{ 2}(x) - 1.5 cos x - 1 = 0

Factor the left side

(cos x - 0.5)(cos x - 2) = 0

cos x - 0.5 = 0 , cos x = 0.5

or cos x - 2 = 0 or cos x = 2 , no real x is solution to this equation.

The given equation has same solutions as the equation

cos x = 1/2

__Questions 24:__

Which point is on the graph of the inverse of the function f(x) = 10^{x + 2}?

A. (100 , 0)

B. (0 , 100)

C. (10 , 0)

D. (0 , 10)

__Solution__

Note that f(0) = 100. Hence

f^{ -1}(100) = 0 , where f^{ -1} is the inverse of f

and therefore the point (100,0) is the graph of the inverse of f

__Questions 25:__

If 10^{x / y} = A / B, then

A. x / y = Log A / Log B

B. y = x / (Log A - Log B)

C. x = y / (Log A + Log B)

D. y = x Log (A / B)

__Solution__

Take the Log of both sides and simplify

Log(10^{x / y}) = Log(A / B)

x / y = Log A - Log B

y / x = 1 / (Log A - Log B)

y = x / (Log A - Log B)

__Questions 26:__

If π < x < 3 π/2, then sin x can be expressed in terms of tan x as follows

A. sin x = tan x / √[ 1 + tan^{2}(x) ]

B. sin x = - √[ 1 + tan^{2}(x) ] / tan x

C. sin x = - tan x / √[ 1 + tan^{2}(x) ]

D. sin x = √[ 1 + tan^{2}(x) ] / tan x

__Solution__

Start with the identity

tan x = sin x / cos x

Square both sides

tan^{2}(x) = sin^{2}(x) / cos^{2}(x)

Use the identity cos^{2}(x) = 1 - sin^{2}(x)

tan^{2}(x) = sin^{2}(x) / (1 - sin^{2}(x))

which gives sin^{2}(x) = tan^{2}(x) / (1 + tan^{2}(x))

Since x is in quadrant III, sin x is negative and therefore

sin x = - √[ tan^{2}(x) / (1 + tan^{2}(x))]

= - tan x / √(1 + tan^{2}(x))

__Questions 27:__

Angle x = 11 π / 3 is coterminal to angle y given by

A. y = π / 3

B. y = π / 6

C. y = 5 π / 3

D. y = 2 π / 3

__Solution__

A possible coterminal angle to x is

11 π / 3 - 2 π = 11 π / 3 - 6 π / 3 = 5 π / 3

__Questions 28:__

If x is such that 3 π/2 < x < 2 π and sin x = -1 / 2, then x =

__Solution__

We first solve sin(y) = 1 / 2

y = π / 6 , which can be used as a reference angle

The solution to sin x = -1 / 2 is equal to

x = 2 π - π/6 = 11 π/6

__Questions 29:__

The exact value of cos(127 π/3) is given by

__Solution__

Let us write 127 π/3 as follows

127 π/3 = 126 π / 3 + π / 3 = 42 π + π / 3

A coterminal angle to 127 π/3 is π/3. Hence

cos(127 π/3) = cos(π/3) = 1/2

__Questions 30:__

If Log(x - y) = 3 and Log(x + y) = 4, then x =

__Solution__

Log(x - y) = 3 is equivalent to

x - y = 10^{3} = 1000

Log(x + y) = 4 is equivalent to

x + y = 10^{4} = 10,000

We now solve the system

x - y = 1000 and x + y = 10,000

2x = 11,000

x = 5,500

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