# Solutions and explanations Maths Questions (2)

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

## Solution to Question 1y = Log(x) if and only ifx = 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 writeLog _{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 ifx ^{ 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 sidex ^{ 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 squarey = 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 atx = - 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 sidesln(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 1825 × π / 180 = 5 π / 36
## Solution to Question 192π / (π/6) = 12
## Solution to Question 20D. For all x real, sin(2x) = 2 sin x cos x
## Solution to Question 21A. 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. Hencetan 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 followscos ^{ 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. Hencef ^{ -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 simplifyLog(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 identitytan 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 is11 π / 3 - 2 π = 11 π / 3 - 6 π / 3 = 5 π / 3
## Solution to Question 28We first solve sin(y) = 1 / 2y = 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 follows127 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 tox - 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 ## More References and Linksmath questions and problems with detailed solutions in this site. |