Cambridge IGCSE And O Level Additional Mathematics 1st Edition Val Hanrahan, Jeanette Powell - Solutions

Solve the following inequalities:i. Graphicallyii. Algebraically.a. |x – 1| < |x + 1|b. |x – 1| > |x + 1|c. |2x – 1| ≤ |2x + 1|d. |2x – 1| ≥ |2x + 1|
Identify these graphs. (They are the moduli of cubic graphs.) a b y A y A 12 12 12 11 11 10 10 10 9- 9- 9 8- 8- 구 구 6, 5- 4 3- 2- -3 -2 -14 1 2 3 4 -5 -4 -3 -2 -1, 1 2 3 -3 -2 -1, 1 2 3 -2 -2 -2 -3 -3 -3 1, 1. 4. 1. 4- 3. 1.
a. Draw the graph of y = |2x + 3|.b. Use the graph to solve the equation |2x + 3| = 7.c. Use algebra to verify your answer to part b.
Each of the following graphs represents an inequality. Name the inequality y. yA 6- 4 b 5- 3. 4 y = [x| 2 4- y =x + 1 2. -3 -2-1, 1 2 3 x -2 -11 1 23 x -2 y = 2 – x -2 -4 -3 -2 -1, 1 2 3 寸 ㅜ 3. 1.
Why is it not possible to identify the following graph without further information? YA -2 -1 3 4
a. Draw the graph of y = |2x − 3|.b. Use the graph to solve the equation |2x − 3| = 7.c. Use algebra to verify your answer to part b.
Solve the equation |x + 1| = |x – 1| both graphically and algebraically.
Solve the equation |x + 5| = |x – 5| both graphically and algebraically.
Solve the equation |2x + 4| = |2x – 4| both graphically and algebraically.
Simplify the following, giving your answers in the form xn:a. 23 × 27b. 5−3 × 54c. 36 ÷ 33d. 65 ÷ 6−4e. (42)3f. (52)−2
Express each of the following as the square root of a single number:a. 3√6b. 5√5c. 12√3d. 10√17
Simplify the following, leaving your answers in standard form:a. (3 × 105) × (2 × 109)b. (2 × 104) × (3 × 10−3)c. (8 × 105) ÷ 103d. (9 × 109) ÷ (3 × 10−3)
Rewrite each of the following as a number raised to a positive integer power:a. 3−2b. 5−4c. (2/3)-3d. (1/3)-6
Simplify the following:a. √25/49b. √24/9c. √12/15d. √6/121
Simplify the following, leaving your answers in standard form:a. (5 × 107) × (3 × 10−3)b. (4 × 10−2) × (6 × 104)c. (4 × 107) ÷ (8 × 10−2)d. (3 × 103) ÷ (6 × 106)
Simplify the following by collecting like terms:a. (3 + √2) + (5 + 4√2)b. 4(√3 − 1) + 4(√3 + 1)
Find the value of each of the following, giving your answer as a whole number or fraction:a. (34 × 3-2)b. 6-6 × 66c. 55 ÷ 52d. (23)4e. 3(32)2f. 7-2g. (1/4)-3h. 2-5i. (3/4)-2j. 91/2k. 811/4l. 163/2m. 272/3n. 256-1/4o. 128-5/7
Rank each set of numbers in order of increasing size:a. 35, 44, 53b. 27, 35, 44c. 2−5, 3−4, 4−3
Expand and simplify:a. (√3 + 2)(√3 − 2)b √3(5 – √3)c. (4 + √2)2d. (√6 – √3)(√6 – √3)
Find the cost of 6 pencils and a ruler. 3 pencils and 4 rulers cost $5.20. 5 pencils and 2 rulers cost $4.
32 = 9 can be written using logarithms as log3 9 = 2. Using your knowledge of indices, find the value of each of the following without using a calculator:a. log2 16b. log3 81 c. log5 125 d. log4 1/64
Rationalise the denominators, giving each answer in its simplest form:a. 1/√6b. 12/√3c. √6/2√2d. 1/(√5 – √2)e. 4 – √2f. 3 – √5/5 + √5
Write the following in the form a + b c where c is an integer and a and b are rational numbers:a. 1 + √2/3 – √2b. 3√5/2 + √5c. 2√6/√6 - 2
Simplify the following:a. 3a2 × 2a5b. 6x4y2 × 2xy−4c. 10b5 ÷ 2b2d. 12p−4q−3 ÷ 3p2q2e. (4m)3f. (2s2t)6
Work out the length of AC. A (3+VZ) cm C (3-V2)cm
Find integers x and y such that 2x × 3y = 64.
A square has sides of length x cm and diagonals of length 12 cm. Use Pythagoras’ theorem to find the exact value of x and work out the area of the square.
An equilateral triangle has sides of length 3 cm. Work out:a. The height of the triangleb. The area of the triangle in its simplest surd form.
Multiply (x3 + 2x2 − 3x − 4) by (x + 1).
Determine whether the following linear functions are factors of the given polynomials:a. x3 − 8x + 7; (x − 1)b. x3 + 8x + 7; (x + 1)c. 2x3 + 3x2 − 4x − 1; (x − 1)d. 2x3 − 3x2 + 4x + 1; (x + 1)
For each function, find the remainder when it is divided by the linear factor shown in brackets:a. x3 + 2x2 − 3x − 4; (x − 2)b. 2x3 + x2 − 3x − 4; (x + 2)c. 3x3 − 3x2 − x − 4; (x − 4)d. 3x3 + 3x2 + x + 4; (x + 4)
Multiply (x3 − 2x2 + 3x + 2) by (x − 1).
Use the factor theorem to find a linear factor of each of the following functions. Then factorise each function as a product of three linear factors and sketch its graph.a. x3 − 7x − 6b. x3 − 7x + 6c. x3 + 5x2 − x − 5d. x3 − 5x2 − x + 5
When f(x) = x3 + ax2 + bx + 10 is divided by (x + 1), there is no remainder.When it is divided by (x − 1), the remainder is 4. Find the values of a and b.
Multiply (2x3 − 3x2 + 5) by (2x − 1).
Factorise each of the following functions completely:a. x3 + x2 + x + 1 b. x3 − x2 + x − 1c. x3 + 3x2 + 3x + 2 d. x3 − 3x2 + 3x − 2
The equation f(x) = x3 + 4x2 + x − 6 has three integer roots.Solve f(x) = 0.
Multiply (x2 + 2x − 3) by (x2 − 2x + 3).
Find the value of x in each of the following:a.b.c.d.e. (1/2)x = 8f. 4x = 1/64g. 2x = 0.125h. 4x = 0.0625 33 x 36 3* 37 +35
For what value of a is (x − 2) a factor of x3 − 2ax + 4?
(x − 2) is a factor of x3 + ax2 + a2x − 14. Find all possible values of a.
Multiply (2x2 − 3x + 4) by (2x2 − 3x − 4).
For what value of c is (2x + 3) a factor of 2x3 + cx2 − 4x − 6?
When x3 + ax + b is divided by (x − 1), the remainder is −12. When it is divided by (x − 2), the remainder is also −12. Find the values of a and b and hence solve the equation x3 + ax + b = 0.
Simplify (x2 − 3x + 2)2.
The expression x3 − 6x2 + ax + b is exactly divisible by (x − 1) and (x − 3).a. Find two simultaneous equations for a and b.b. Hence find the values of a and b.
Sketch each curve by first finding its points of intersection with the axes:a. y = x3 + 2x2 − x − 2b. y = x3 − 4x2 + x + 6c. y = 4x − x3d. y = 2 + 5x + x2 − 2x3
Divide (x3 − 3x2 + x + 1) by (x − 1).
Divide (x3 − 3x2 + x + 2) by (x − 2).
Divide (x4 − 1) by (x + 1).
Divide (x2 − 16) by (x + 2).
Solve the following pairs of simultaneous equations graphically:a. y = x + 2y. = 2x − 3b. x + 2y = 32x - y = -4
Solve this pair of simultaneous equations algebraically:x2 + y2 = 13x = 2
Use the substitution method to solve the simultaneous equations in the following Question.a. 2x + y = 13y = 2x + 1b. x + 2y = 13x = 2y + 1
Solve this pair of simultaneous equations algebraically and sketch a graph to illustrate your solution:y = x2 − x + 8y = 5x
Use the substitution method to solve the simultaneous equations in the following Question.a. 3x + 4y = 2y = 4x + 10b. 4x + 3y = 2x = 4y + 10
Solve this pair of simultaneous equations:xy = 4y = x − 3
Use the substitution method to solve the simultaneous equations in the following Question.x - 3y = -2y = 3x − 2
Solve this pair of simultaneous equations:y = 8x2 − 2x − 104x + y = 5
Use the substitution method to solve the simultaneous equations in the following Question.x + 4y = −13x = 3y + 1
The diagram shows the circle x2 + y2 = 25 and the line x − 7y + 25 = 0. Find the coordinates of A and B. y 6- x- 7y + 25 = 0 up B 4 A 3- 2- -8 -7 -6 -5 -4 -3 -2 -1, 1 2 3 4 $ 6 7 8 -2 - -3 - -4 x2 + y2 = 25 -6
Use the elimination method to solve the simultaneous equations in the following question.a. x + y = 4x - y = 2b. x + 2y = 4x - 2y = 2
The diagram shows a circular piece of card of radius r cm, from which a smaller circle of radius x cm has been removed. The area of the remaining card is 209πcm2. The circumferences of the two circles add up to 38π. Write this information as a pair of simultaneous equations and hence find the
Use the elimination method to solve the simultaneous equations in the following question.a. 3x + y = 92x - y = 1b. 3x + 2y = 9x - y = 0.5
a. Solve this pair of simultaneous equations:y = x2 + 1y = 2xb. Why is there only one solution? Illustrate this using a sketch.
Use the elimination method to solve the simultaneous equations in the following question.2x + 3y = −44x + 2y = 0
a. Explain what happens when you try to solve this pair of simultaneous equations:y = 2x2 − 3x + 4y = x − 1b. Illustrate your explanation with a sketch graph.
Use the elimination method to solve the simultaneous equations in the following question.5x - 2y = -233x + y = −5
At the cinema, 3 packets of popcorn and 2 packets of nuts cost $16 and 2 packets of popcorn and 1 packet of nuts cost $9. What is the cost of one packet of each?
Two adults and one child paid $180 to go to the theatre and one adult and three children paid $190. What it is the cost for two adults and five children?
A shop is trying to reduce their stock of books by holding a sale. $20 will buy either 8 paperback and 4 hardback books or 4 paperbacks and 7 hardbacks. How much change would I get from $40 if I bought 10 paperbacks and 10 hardbacks?
By first writing each of the following equations using powers, find the value of y without using a calculator:a. y = log2 8b. y = log3 1c. y = log5 25d. y = log2 1/4
For each set of graphs:i. Sketch the graphs on the same axes.ii. Give the coordinates of any points of intersection with the axes.a. y = ex, y = ex + 1 and y = ex+1b. y = ex, y = 2ex and y = e2xc. y = ex, y = ex − 3 and y = ex−3
Sketch the graphs of y = e3x and y = e3x − 2.
Find the following without using a calculator:a. lg100b. lg(one million)c. lg 1/1000d. lg(0.000 001)
Sketch the graphs of y = e2x, y = 3e2x and y = 3e2x − 1.
Using the rules for manipulating logarithms, rewrite each of the following as a single logarithm. For example, log 6 + log 2 = log(6 × 2) = log 12.a. log3 + log5b. 3log4c log12 − log3d. 1/2 log25e. 2log3 + 3log2f. 4log3 − 3log4g. 1/2 log4 + 4log 1/2
Sketch each curve and give the coordinates of any points where it cuts the y-axis.a. y = 2 + exb. y = 2 − exc. y = 2 + e−xd. y = 2 − e−x
Express each of the following in terms of log x:a. log x5 − log x2b. log x3 + 3log xc. 5log√x − 3log 3√x
Solve the following equations:a. 5e0.3t = 65b. 13e0.5t = 65c. et+2 = 10d. et−2 = 10
This cube has a volume of 800 cm3.a. Use logarithms to calculate the side length correct to the nearest millimetre.b. What is the surface area of the cube?
The value, $V, of an investment after t years is given by the formula V = Ae0.03t , where $A is the initial investment.a. How much, to the nearest dollar, will an investment of $4000 be worth after 3 years?b. To the nearest year, how long will I need to keep an investment for it to double in value?
Starting with the graph of y = ln x, list the transformations required, in order when more than one is needed, to sketch each of the graphs. Use the transformations you have listed to sketch each graph.a. y = 3ln xb. y = ln(x + 3)c. y = 3ln2xd. y = 3ln x + 2e. y = −3ln(x + 1)f. y = ln(2x + 4)
The path of a projectile launched from an aircraft is given by the equation h = 5000 − e0.2t, where h is the height in metres and t is the time in seconds.a. From what height was the projectile launched? The projectile is aimed at a target at ground level.b. How long does it take to reach the
Match each equation from i to vi with the correct graph a to f.i. y = log(x + 1)ii. y = log(x − 1)iii. y = −ln xiv. y = 3ln xv. y = log(2 − x)vi. y = ln(x + 2) a 6- 5- 4. 1- 2 1. -2 -1 /1 e 3 4 -1 1- -1 3 -1 -1- LO 2. 2. m. -- 4. 3. 2. 2.
Match each equation from i to vi to the correct graph a to f.i. y = e2xii. y = ex + 2iii. y = 2 − exiv y = 2 − e−xv. y = 3e−x − 5vi. y = e−2x − 1 6- 5- 4- 5- 3- 4- 1. -2- -2 -1 4- 4- 3- 2- -3 -2 -1 -2 -1 4, 4. 3. w. N. 2. m. 2.
Solve the following equations for x, given that lna = 3:a. a2x = e3b. a3x = e2c. a2x − 3ax + 2 = 0
A radioactive substance of mass 100 g is decaying such that after t days the amount remaining, M, is given by the equation M = 100e−0.002t.a. Sketch the graph of M against t.b. What is the half-life of the substance (i.e. the time taken to decay to half the initial mass)?
Before photocopiers were commonplace, school examination papers were duplicated using a process where each copy produced was only c% as clear as the previous copy. The copy was not acceptable if the writing was less than 50% as clear as the original. What is the value of c if the machine could
When David started his first job, he earned $15 per hour and was promised an annual increment (compounded) of 3.5%.a. What is his hourly rate be in his 5th year?After 5 years he was promoted. His hourly wage increased to $26 per hour, with the same compounded annual increment.b. For how many more
Use logarithms to solve the equation 52x−1 = 4x+3. Give the value of x correct to 3 s.f.
a. Solve 2(32x) − 5(3x) + 2 = 0b. Solve exex+1 = 10c. Solve 22x − 5(2x) + 4 = 0
a. $20 000 is invested in an account that pays interest at 2.4% per annum. The interest is added at the end of each year. After how many years will the value of the account first be greater than $25 000?b. What percentage interest should be added each month if interest is to be accrued monthly?c.
For each of the following pairs of points A and B, calculate:i. The gradient of the line ABii. The gradient of the line perpendicular to ABiii. The length of ABiv. The coordinates of the midpoint of AB.a. A(4, 3) B(8, 11)b. A(5, 3) B(10, –8)c. A(6, 0) B(8, 15)d. A(–3, –6) B(2, –7)
Match the equivalent relationships.i. y = prxii. y = rpxiii. y = pxriv. y = xpra. log y = log p + r log x b. log y = logr + xlog pc. lg y = lg p + xlgrd. lg y = lg x + r lg p
A(0, 5), B(4, 1) and C(2, 7) are the vertices of a triangle. Show that the triangle is right angled:a. By working out the gradients of the sidesb. By calculating the lengths of the sides.
For each of the following models, k, a and b are constants. Use logarithms to base e to rewrite them in the form y = mx + c, stating the expressions equal to x, y, m and c in each case.a. y = kaxb. y = kxac. y = akxd. y = axk
A(3, 5), B(3, 11) and C(6, 2) are the vertices of a triangle.a. Work out the perimeter of the triangle.b. Sketch the triangle and work out its area using AB as the base.
The table below shows the area, A, in square centimetres of a patch of mould t days after it first appears.It is thought that the relationship between A and t is of the form A = kbt.a. Show that the model can be written as logA = (logb)t + logk.b. Plot the graph of logA against t and explain why it
A quadrilateral PQRS has vertices at P(–2, –5), Q(11, – 7), R(9, 6) and S(–4, 8).a. Work out the lengths of the four sides of PQRS.b. Find the coordinates of the midpoints of the diagonals PR and QS.c. Without drawing a diagram, show that PQRS cannot be a square. What shape is PQRS?

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Cambridge IGCSE And O Level Additional Mathematics 1st Edition Val Hanrahan, Jeanette Powell - Solutions

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