Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions by Ibrahim Wazir, Tim Garry (2019)

Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions by Ibrahim Wazir, Tim Garry (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions by Ibrahim Wazir, Tim Garry (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions by Ibrahim Wazir, Tim Garry (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions by Ibrahim Wazir, Tim Garry (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions by Ibrahim Wazir, Tim Garry (2019)

WORKED SOLUTIONS

(h) Factor out k at the denominator of the fraction:

2 In parts (a) to (d), find m, the slope of the line, using , then find the y-intercept c,

by choosing one of the given points and substituting its coordinates and the value of the slope into the slope-intercept form of the equation of a line,

(a)

(b) EITHER:

OR: The given points have the same y-coordinate, so the line is horizontal,

the equation of the line is

f

gh k gh fk k

h g f

(c)

(d) By inspection, the given points have the same x-coordinate, so the line is vertical,

and its equation is (e)

Parallel lines have the same slope, consequently the slope of the required line

is

(f)

The slope of a line perpendicular to another line with slope is

3 In each of these, substitute the coordinates of the given points in the relevant formulae:

(a)

(b)

(c)

(d)

4 Substitute the coordinates of the given points in the distance formula and set it equal to 5:

(a)

(b)

5 Use the distance formula to show that properties regarding the length of sides for the indicated

shapes hold true

(a) Let , and be the vertices of triangle

The lengths of the sides are:

is a right-angled triangle

is a right-angled triangle

(b) Let , and be the vertices of triangle

is isosceles, because two of its sides have the same length (c) Let , , and be the vertices of quadrilateral

If and , then is a parallelogram,

is a parallelogram

6 (a) Subtract the two equations to eliminate x:

Substitute into either of the original equations to find x:

The solution is (b) Multiply the second equation by 3 and add it to the first equation to eliminate y:

Substitute into either of the original equations to find x:

The solution is (c) Multiply the first equation by 4 and the second equation by 3, then subtract the two

equations, to eliminate y:

Substitute into either of the original equations to find x:

The solution is (d) Multiply the second equation by 4 and add it to the first equation to eliminate x:

This is not true, so the system of equations has no solution

(e) Multiply the first equation by 5 and the second equation by 7,

then add the two equations, to eliminate y:

Substitute into either of the original equations to find y:

The solution is

7 (a) Make y the subject in the first equation:

Substitute the expression of y in the second equation and solve for x:

Substitute into either of the original equations to find y:

The solution is (b) Make y the subject in the second equation:

Substitute the expression of y in the first equation and solve for x:

Substitute into the expression of y:

The solution is (c) Divide the first equation by 2:

Make x the subject in this equation:

Substitute the expression of x in the second equation and solve for y:

true for all This means that there are an infinite number of solutions, due to the fact that the two equations are multiples of each other (the lines representing the two equations are coincident) It follows that the solution of this system is the set of all points on the line

(d) Make y the subject in the second equation:

Substitute the expression of y in the first equation and solve for x:

Substitute into the expression of y:

The solution is (e) Make y the subject in the second equation:

Substitute the expression of y in the first equation and solve for x:

Substitute into the expression of y:

The solution is (f) Multiply the second equation by 10:

Make y the subject in this equation:

Substitute the expression of y in the first equation and solve for x:

Substitute into the expression of y:

The solution is

8 (a) The solution is

The solution is

The solution is (c)

infinite solutions (d)

The solution is

10 (a) The solution is

(b) The system of equations has no solution

(c) The solution is

(d) Infinite solutions

11

The last equation is:

In order for the system to have no solutions, the coefficient of z must be 0,

and the right side of the equation should not be equal to 0

When , the right-hand side of the equation is:

the required value is

Exercise 1.2

1 (a) (i) Graph G (ii) represents a straight line with a positive slope

, passing through the origin (b) (i) Graph L (ii) represents a horizontal line

(c) (i) Graph H (ii) represents a straight line

with positive slope , and with a negative y-intercept

centred at the origin and with radius 2 (this is not a function)

(e) (i) Graph J (ii) represents a straight line with a negative

slope , and with a positive y-intercept

(f) (i) Graph C (ii) represents a parabola opening upwards,

with vertex at (g) (i) Graph A (ii) (h) (i) Graph I (ii) represents a rational function with asymptotes

and

(i) (i) Graph F (ii) represents a parabola

opening downwards, with vertex at

3 The area of a triangle is , where b and h are the base and the height of the triangle,

respectively Let l be the side of the equilateral triangle

When drawing one of the heights, a right-angled triangle is formed with sides h, l and

4 The area of the pavement can be calculated as the difference

between the areas of two rectangles:

8 (a) The domain is the set of all x-values:

(b) The radius is a length, so

(c) f is a linear function, so

(d) h is a quadratic function, so

(e) The radicand must be greater or equal to zero, so

(f) All real numbers can be cube rooted, so

(g) f is a rational function; its denominator cannot be equal to 0

2 l

(h) There are two conditions: x cannot be 0, as it is at the denominator of a fraction, and

, consequently the domain is ,

9 No, because a line with equation , where c is a constant, fails the requirement of a

function to map one x-value to only one y-value

10 (a) Substitute the given x-value into the expression of the function:

(i)

(iii) (b) The radicand must be negative for the function to be undefined,

so (c) The domain of h is given by , the range is

(the square root of any positive number is positive, the square root of 0 is 0)

11 (a) (i) is a rational function, its denominator cannot be equal to 0,

so the domain is The range of f is

, as the function can take all real values, except 0,

(b) (i) is a rational function with a square root at the denominator,

so the radicand has to be strictly positive, so meaning the domain is The values given by function g will be strictly positive, because the square root at the denominator will always result in a positive number, so the range of g is

(ii) To find the x-intercept, set y (or ) equal to 0, and solve for x:

no solution, so no x-intercept

The y-intercept is found when , but this value for x is not in the domain

of the function, so there is no y-intercept

there are two vertical asymptotes:

the equation of the horizontal asymptote is The graph of the function is:

(c) (i) is a rational function, its denominator cannot be equal to 0,

(d) (i) is a square root function,

so the radicand must be positive or 0:

the domain is

The values given by function g will be positive, because the square root at the denominator will always result in 0 or a positive number, but, because the radicand is a quadratic expression, the range of must also be considered The x-coordinate of the vertex of the downwards parabola representing is consequently the y-coordinate ismeaning the y-values given by g are always less than 5, This means that the range of p is

(ii) To find the x-intercept, set y (or ) equal to 0, and solve for x:

the x-intercepts are and

To find the y-intercept, set x equal to 0, and calculate y:

the y-intercept is This function does not have any asymptotes

The graph of the function is:

(e) (i) is a rational function, its denominator cannot be equal to 0,

, this means that the range of f is (ii) To find the x-intercept, set y (or ) equal to 0, and solve for x:

the x-intercept is The y-intercept is found when , but this value for x is not in the domain

of the function, so there is no y-intercept

The equation of the vertical asymptote is , the equation of the horizontal

(g)

(h)

(i)

(j)

3 In each question, when finding the domain of , check the following two conditions:

the input x is in the domain of , because the first rule to be applied to x is g,

the inside function of the output is in the domain of f (the range of g must be either equal to

or a subset of the domain of f )

The same applies when finding the domain of : x must be in the domain of , because the first rule to be applied to x is f, the inside function of , and is in the domain of g (a)

, this set of values is the same as the domain for g, so the domain

of is also

(b)

, this set of values is the same set of values as the domain for f,

so the domain of is also

Domain Range

Domain Range

, this set of values is a subset of the domain of g, so the domain

of is also (c)

, this set of values is a subset of the domain of f, so the domain

of is also

, this set of values is a subset of the domain of g, so the

domain of is also (d)

, but this set of values is not a subset of the domain of f To be able to compose the two functions, the x-value, which is the input of g resulting in an output of , must be excluded from the domain of g

Domain Range

Domain Range

, this set of values is the same as the domain for f, so the domain of

is also

, this set of values is the same as the domain for g, so the domain

of is also (f)

, this set of values is a subset of the domain of f, so the domain

of is also

, this set of values is the same as the domain for g, so the domain

of is also (g)

, but this set of values is not a subset of the domain of f To be able to compose the two functions, the x-value, which is the input of g resulting in an output of must be excluded from the domain of g

Domain Range

Domain Range

, but this is not a subset of the domain of g To be able to

compose the two functions, the x-value, which is the input of f resulting in an output

of must be excluded from the domain of f

(h)

, this set of values is the same as the domain for f,

so the domain of is also

, this set of values is the same as the domain for g,

so the domain of is also (i)

, but this set of values is not a subset of the domain of f To be able to compose the two functions, the x-value, which is the input of g resulting in an output of must be excluded from the domain of g

, so the domain of is

, this set of values is a subset of the domain of g,

so the domain of is also

Domain Range

Domain Range

4 (a)

, this is not a subset of the domain of f To be able to compose the two functions, only the x-values which are common to both the domain of g (real numbers greater than or equal to 1) and the range of h (real numbers less than or equal

to 10) are acceptable, so only real numbers between 1 and 10, including 1 and 10, can

be inputs for g This means that the domain of will not contain the x-values that will lead to outputs which are less than 1

, so the domain of is: The outputs of , corresponding to inputs taking values from the set

, will be elements of the set , so the range of is

(b)

, this is a subset of the domain of h, so the domain of

is also The range of is , as the restriction on the domain of h does not impact the range of the quadratic

5 (a)

, this set of values is not a subset of the domain of f To be able to compose the two functions, the x-value, which is the input of g resulting in an output of must be excluded from the domain of g

, so the domain of is:

The outputs of will be elements of the set , as the input of g cannot be 0, so the range of is:

(b)

Domain Range

Domain Range

, this set of values is a subset of the domain of g, so the domain

The outputs of , corresponding to inputs taking values from the set

, will be elements of the set , ( is obtained for , which is not in the domain), so the range of is:

6 In each of the following, try to identify the rules transforming the input into the expression

given by , and take into consideration the order in which the two functions are combined.(a) 3 is added to x, and the result is squared

(b) 5 is subtracted from x, and the result is square rooted

(c) x is square rooted, and the result is subtracted from 7

(d) 3 is added to x, and the reciprocal of the result is computed

(e) 1 is added to x, this result is then the power to which 10 is raised (f) 9 is subtracted from x, and the result is cube rooted

(g) 9 is subtracted from the square of x, and the absolute value of

result is taken(h) the square root of the difference between x and 5, and the reciprocal

of the result is computed

, but this set of values is not a subset of the domain of f To be able to compose the two functions, the x-values, which are the inputs of g resulting in the outputs must be excluded from the domain of f

,

so the domain of is

(d)

, this set of values is the same set of values as the domain for f,

so the domain of is also

Domain Range

Domain Range

Domain Range

Exercise 1.4

3 If then so is an output for the original function f

The two graphs are reflections of each other in the line , so f and g are inverse functions.

6 The fact that the domain of is the same as the range of is used when answering the

following questions

(a)

The domain of is

(b)

The domain of is (c)

The domain of is (d)

The domain of is (e)

The domain of is

(f)

The domain of is (g)

The domain of is (h)

The domain of is

(i)

The domain of is (j)

The domain of is

7 (a) The graph of shows that f is a decreasing function on its

domain (no need for restricting the domain) This means that f is a one-to-one function, consequently exists

Next, find the expression of :

The domain of is the range of f , namely The two graphs are shown below:

(b) The graph of shows that f is not a one-to-one function on its domain In order for to exist, the domain of f must be restricted to either or

Next, find the expression of for :

The domain of is the range of f , namely The graphs are shown below for a domain for f restricted to :

(c) The graph of shows that f is not a one-to-one function on its

domain In order for to exist, the domain of f must be restricted to either

or

Next, find the expression of for :

The domain of is the range of f , namely The graphs are shown below for a domain for f restricted to :

(d) The graph of shows that f is not a one-to-one function on its

domain In order for to exist, the domain of f must be restricted to either

or Next, find the expression of for :

The domain of is the range of f , namely The graphs are shown below for a domain for f restricted to :

(g)

(h)

10

Exercise 1.5

1 (a) The graph of is a vertical translation of the parabola representing the

basic function The transformed graph is obtained by shifting all points on theoriginal parabola 6 units down, so the new vertex will be at , and the

new x-intercepts are at and

(b) The graph of is a horizontal translation of the basic function

All points on the original parabola will shift 6 units right, including the original vertex, , which will now be located at

(c) The graph of is a vertical translation of the graph representing the

basic function The transformed graph is obtained by shifting all points on theoriginal graph 4 units up, including the original x-intercept, , now at

(d) The graph of is a horizontal translation of the graph representing the

basic function The transformed graph is obtained by shifting all points on theoriginal graph 4 units left, including the original x-intercept, , now at

(e) The graph of is obtained by translating all points on the graph of

the basic function by 2 units right and 5 units up The original x-intercept, , will now be located at

(f) The graph of is a horizontal translation of the basic function

All points on the original graph will shift 3 units right, including the original vertical asymptote, , which will now have equation The horizontal asymptote is the same,

(g) The graph of is a translation of the graph of the basic function

All points on the original graph will shift 5 units left and 2 units up, including the original asymptotes: the equations of the new vertical and horizontal asymptotes are and , respectively

(h) The graph of is obtained by transforming the graph of the basic

function First, the original graph is reflected in the x-axis, then all points shift

4 units down, including the original x-intercept, which is now at

(i) The graph of is obtained by transforming the graph representing

the basic function First, shift all points on the original graph 1 unit right, then reflect the graph in the x-axis, and shift it 6 units up The original x-intercept, , will first shift to , then to The reflection in the x-axis has no effect on point , as it is on the line of reflection

(j) The graph of is obtained by transforming the graph of the basic

function as follows: first, shift all points on the graph 3 units left, then reflect

it in the x-axis The transformation can also be performed in a different order if we consider the equivalent form of the equation of f , : first, reflect the graph in the x-axis and then shift it right 3 units, you will obtain the same graph as before In both cases, the original x-intercept, , will end up at

(k) The graph of is obtained by transforming the graph of the basic

function by stretching it vertically using a scale factor of 3 (the y-coordinates

of all points on the original graph are multiplied by 3) The x-intercept, , is not impacted by this transformation, it stays at

(l) The graph of is obtained by transforming the graph of the basic quadratic

function by stretching it vertically using a scale factor of (the y-coordinates

of all points on the original graph are multiplied by ) The x-intercept, , is not impacted by this transformation, it stays at

(m) The graph of is obtained by transforming the graph of the basic

quadratic function by stretching it horizontally using a scale factor of 2 (the x-coordinates of all points on the original graph are multiplied by 2)

The transformation can also be performed in a different way if we consider the equivalent form of the equation of f , , this can be done by vertically stretching the graph of with a scale factor of (the y-coordinates of all points on the original graph are multiplied by ) In both cases, the original x- intercept, , will not be impacted by the transformation, it stays at

(n) The graph of is obtained by transforming the graph of the basic

function by reflecting the original graph in the y-axis Alternatively, if we consider the equivalent form of the equation of f, , then the graph of

should be reflected in the x-axis In both cases, the original x-intercept, , will not be impacted by the transformation; it stays at

2 (a) The given graph represents a parabola, so the transformations are applied to the graph

of the basic quadratic function Based on the given graph, the transformations are: reflection in the x-axis, followed by a vertical shift of 5 units up

the equation of the graph is (b) The shape of the given graph suggests that the basic function to work with is

The graph is a reflection in the y-axis, as checked when considering the point This is the image of point , so the required equation is

(c) The basic graph in this case is , the applied transformations are: reflection in

the x-axis, followed by a horizontal shift of 1 unit left

the equation of the graph is (d) The graph to be transformed is , the applied transformations are: a horizontal

translation of 2 units right, followed by a vertical shift of 3 units down (this is easily deduced when looking at the original vertical and horizontal asymptotes, the lines and have changed to and , respectively)

the equation of the graph is

3 In the following questions, first recognise the transformations to be applied by analysing the

given equation, then transform the important points of the original graph: the two ends,

and , the maximum point, , and the minimum point, , then plot and join these images to obtain the final graph

(a) The transformation to be applied is a vertical shift of 3 units down

,

(b) The transformation to be applied is a horizontal shift of 3 units to the right