
Solution to Question 1y = Log(x) if and only if
x = 10^{y}
Interchange x and y
y = 10^{x}
Hence
f^{ 1}(x) = 10^{x}
Solution to Question 2Use 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 3Rewrite 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
Solution to Question 4Log_{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
Solution to Question 5Expand 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 6Write 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 7Checking 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 8We 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 9We 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 10The 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 11The 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 12Take 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 13How 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 14Let 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
Solution to Question 15Select 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 16B is not true.
tan ( x) =  tan x , odd function
Solution to Question 17D 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
Solution to Question 19
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 22tan 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 23Let'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 24Note 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 25Take 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 26Start 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))
Solution to Question 27A possible coterminal angle to x is
11 π / 3  2 π = 11 π / 3  6 π / 3 = 5 π / 3
Solution to Question 28We 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 29Let 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 30Log(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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