# Solutions and explanations Maths Questions (2)

Solutions and explanations to the math questions (2) in this site are presented.

## Solution to Question 1

y = Log(x) if and only if
x = 10
y
Interchange x and y
y = 10
x
Hence
f
-1(x) = 10x

## Solution to Question 2

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)

## Solution to Question 3

Rewrite equation using 16 = 24
2
3x - 1 = 24
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

## Solution to Question 4

Logx 9 = 2 if and only if
x
2 = 9
x
2 = 32
For the above exponential expressions to be equal, their bases must be equal. Hence
x = 3

## Solution to Question 5

Expand the polynomials 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

## Solution to Question 6

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)

## Solution to Question 7

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

## Solution to Question 8

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.

## Solution to Question 9

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]

## Solution to Question 10

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)

## Solution to Question 11

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)

## Solution to Question 12

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)

## Solution to Question 13

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�

## Solution to Question 14

Let r be the distance from the (0,0) to (-3,4).
r = √((-3)
2 + 42) = 5
sec x = 1/cos x = 1 / (-3/5) = -5/3

## Solution to Question 15

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)

## Solution to Question 16

B is not true.
tan (- x) = - tan x , odd function

## Solution to Question 17

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

## Solution to Question 18

25 � π / 180 = 5 π / 36

2π / (π/6) = 12

## Solution to Question 20

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

## Solution to Question 21

A. is not an identity.

## Solution to Question 22

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

## Solution to Question 23

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

## Solution to Question 24

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

## Solution to Question 25

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)

## Solution to Question 26

tan x = sin x / cos x
Square both sides
tan
2(x) = sin2(x) / cos2(x)
Use the identity cos2(x) = 1 - sin2(x)
tan
2(x) = sin2(x) / (1 - sin2(x))
which gives sin
2(x) = tan2(x) / (1 + tan2(x))
Since x is in quadrant III, sin x is negative and therefore
sin x = - √[ tan
2(x) / (1 + tan2(x))]
= - tan x / √(1 + tan
2(x))

## Solution to Question 27

A possible coterminal angle to x is
11 π / 3 - 2 π = 11 π / 3 - 6 π / 3 = 5 π / 3

## Solution to Question 28

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

## Solution to Question 29

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

## Solution to Question 30

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