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Core 3 examples
Fichas del conjunto (50)
- 1Edexcel A-level - MathematicsAlgebraic Fractions
- 2Example 1 Find the simplest forms of the functions a) 16/20 b) {x+3} / {2x+6} c) {x+2} / {3x+8}a) = 16/20
= {4*4} / {5*4}
=4/5
b) = {x+3} / {2x+6}
= {x+3} / {2(x+3)}
= 1/2
c) = {x+2} / {3x+8}
No simplification possible - 3Example 2 Simplify {1/2x +1} / {1/3x +2/3}= {(1/2x +1)*6} / {(2/3x +2/3)*6}
= {3x+6} / {2x+4}
= {3(x+2)} / {2(x+2)}
= 3/2 - 4Example 3 Simplify the expressions: a) {x^2 -1} / {x^2 +4x +3} b) {x - (1/x)} / {x+1}a) = {x^2 -1}/{x^2 +4x +3}
= {(x+1)(x-1)} / {(x+1)(x+3)}
= {x-1}/{x+3}
b) = {(x- (1/x))*x}/{(x+1)*x}
= {x^2 +1} / {x^2 +x}
= {(x-1)(x+1)}/{x(x+1)}
= {x-1} / x - 5Example 4 Calculate: a) (1/2)*(3/5) b) (a/b)*(c/d)a) = (1/2)*(3/5)
= {1*3}/{2*5}
= 3/10
b) = (a/b)*(c/d)
= {a*c}/{b*d}
= {ac}/{bd} - 6Example 5 Simplify the following products: a) (3/5)*(5/9) b) (a/b)*(c/a) c) ({x+1}/2)*(3/{x^2 -1})a) = (3/5)*(5/9)
= {3*5}/{5*9}
= 15/45
=1/3
b) = (a/b)*(c/a)
= {a*c}/{b*d}
= c/d
c) = ({x+1}/2)*(3/{x^2 -1})
=({x+1}/2)*(3/{(x+1)(x-1)})
=(1/2)*(3/{x-1})
= 3/{2(x-1)} - 7Example 6 Divide 5/6 by 1/3= (5/6)/(1/3)
= (5/6)*(3/1)
= (5/2)*(1/1)
= 5/2 - 8Example 7 Simplify: a) (a/b)/(a/c) b) ({x+2}/{x+4}) / ({3x+6}/{x^2 -16})a) = (a/b)/(a/c)
= (a/b)*(c/a)
= (1/b)*(c/1)
= c/b
b) = ({x+2}/{x+4}) / ({3x+6}/{x^2 -16})
= ({x+2}/{x+4}) * ({x^2 -16}/{3x+6})
= ({x+2}/{x+4}) * ({(x+4)(x-4)}/{3(x+2)})
=(1/1) * ({x-4}/3)
= {x-4}/3 - 9Example 8 Add 1/3 to 3/4= (1/3)+(3/4)
= (4/12) + (9/12)
= 13/12 - 10Example 9 Add 1/3 to 5/12= (1/3) + (5/12)
= (4/12) + (5/12)
= 9/12
=3/4 - 11Example 10 Simplify the following fractions: a) (a/x)+b b) (3/{x+1}) / (4x/{x^2 -1})a) = (a/x)+b
= (a/x) + (b/1)
= (a/x) + (bx/x)
= ({a + bx}/x)
b) = (3/{x+1}) / (4x/{x^2 -1})
= (3/{x+1})*({x^2 -1}/4x)
= (3/{x+1})*({(x+1)(x-1)}/4x)
= (3/1)*({x-1}/4x)
= {3(x-1)}/4x - 12Example 11 Divide (x^3)+(x^2)-7 by (x-3) by using long divisionLook at textbook for the method
= {(x^3)+(x^2)-7}/(x-3)
= x^2 +4x+12 remainder 29 - 13Example 12 Divide x^3 + x^2 -7 by x-3 by using the remainder theoremf(x) = x^3 +x^2 -7
x-3=0
x=3
f(3) = (3)^3 +(3)^2 -7
=29
Remainer is 29
Look at textbook for the method
Answer = x^2 +4x+12 r29 - 14Example 13 Divide (x^4 +x^3 +x -10) by (x^2 +2x -3)Look at textbook for the method
= (x^4 +x^3 +x -10)/(x^2 +2x -3)
=x^2 -x +5 r-12x+5 - 15FunctionsAll examples in functions from the Edexcel Core 3 book
- 16Example 1 Draw a mapping diagram and grphs for the following operations: a) 'add 3' on the set (-3, 1, 4, 6, x) b) 'square' on the set (-1, 1, -2, 2, x)Look at the textbook for the diagram
a) Set A= -3, 1, 4, 6, x
Set B = 0, 4, 7, 9, x+3
b) Set A= -1, 1, -2, 2, x
Set B, 1, 4, x^2 - 17Example 2 Given that the function g(x) = 2x^2 +3, find: a) the value of g(2) b) the value of a such that g(a) = 15 c) the range of the function g(x)a) g(2) = 2(2)^2 +3 = 11
b) g(a) = 2a^2 +3 =36
2a^2 = 32
a^2 = 16
a = +/-4
c) Look at graph in the textbook
Range of g(x) is >= 3 - 18Example 3 (Part 1) Which of the following mappings represent functions? Give reasons for your answer a) Set A = 9, 4, 1 Set B = 3, -3, 2, -2, 1, -1 b) Look at the textbook for the graph mappinga) Every element of set A gets mapped to two elements in set B. Therefore this isn't a function. It is an operation (Square root).
b) Every value of x gets mapped to one value of y. This, therefore, is a function. It can be written in the form y = ax + b for some constants a and b. This type of function is called one-to-one because every element of the range comes exactly from one element in the domain - 19Example 3 (Part 2) Which of the following mapping represent functions? Give reasons for your answers. c) Look at the textbook for the graph mapping d) Look at the textbook for the graph mappingc) On the sketch of y = 1/x, you can see that x = 0 doesn't get mapped anywhere. Therefore not all elements of set A get mapped to elements in set B. Hence it is not a function.
d) This is a function. It is called a many-to-one function because two elements of the domain get mapped to one element in the range - 20Example 4 Find the range of the following functions: a) f(x) = 3x - 2, domain {x = 1, 2, 3, 4} b) g(x) = x^2, domain {xER, -5 =< x =< 5} c) h(x) = 1/x, domain {xER, 0 < x =< 3} State if the functions are one-to-one or many-to-onea) f(x) = 3x - 2, {x = 1, 2, 3, 4}
Set A= 1, 2, 3, 4
Set B= 1, 4, 7, 12
Range of f(x) is {1, 4, 7, 10}, f(x) is one-to-one
b) g(x) = x^2, {-5 =< x =< 5}
Look at graph in the textbook
Range of g(x) is 0 =< g(x) =< 25, g(x) is many-to-one - 21Example 5 The function f(x) is defined by f(x) = {5 - 2x, x < 1 ={ x^2 +3, x >= 1 a) Sketch f(x) stating its range b) Find the values of a such that f(a) = 19a) f(1), using f(x) = 5 - 2x = 5 - 2(1) = 5 - 2 = 3
f(1), using x^2 +3 = (1)^2 +3 = 1 + 3 = 4
Look at the graph in the textbook
The range is the values that y takes and therefore f(x) > 3
b) Look at the graph in the textbook
The positive point is where
x^2 +3 = 19
x^2 = 16
x = 4
The negative pointis where
5 -2x = 19
-2x = 14
x = -7
The two values of a are 4 and -7 - 22Example 6 Given f(x) = x^2 and g(x) = x + 1, find: a) fg(1) b) fg(3) c) fg(x)a) fg(1) = f(1 + 1)
=2^2
=4
b) fg(3) = f(3 + 1)
=4^2
=16
c) fg(x) = f(x + 1)
=(x + 1)^2 - 23Example 7 The functions f and g are defined by f(x) = 3x + 2 and g(x) = x^2 + 4, Find: a) the function fg(x) b) the function gf(x) c) the function f^2 (x) d) the vakues of b such that fg(b) = 62a) fg(x) = f(x^2 + 4
=3(x^2 + 4) + 2
= 3x^2 + 14
b) gf(x) = f(3x + 2)
=(3x + 2)^2 +4
=9x^2 +12x + 8
c) f^2 (x) = f(3x + 2)
= 3(3x + 2) + 2
= 9x + 8
d) fg(x) = 3x^2 + 14
If fg(b) = 62
3b^2 + 14 = 62
3b^2 = 48
b^2 = 16
b = +/- 4 - 24Example 8 The functions m(x), n(x) and p(x) are defined by m(x) = 1/x, n(x) = 2x + 4, p(x) = x^2 - 2. Find in terms of m, n and p the functions a) 2/x + 4 b) 4x^2 +16x +14 c) 1/{4x + 12}a) (2/x) + 4 = 2(1/x) + 4
=2m(x) + 4
=nm(x)
b) 4x^2 +16x +14 = (2x + 4)^2 - 2
= [n(x)]^2 - 2
= pn(x)
c) 1/{4x + 12} = 1/{2(2x + 4) + 4
= 1/{2n(x) + 4}
=1/{nn(x)}
=mnn(x)
= m n^2 (x) - 25Example 9 Find the inverse of the function h(x) = 2x^2 - 7Look at the textbook for the flowchart
h(x) = 2x^2 - 7
Let y = h(x)
y = 2x^2 - 7
y + 7 = 2x^2
¬{{y + 7}/2}=x
Let x = h^-1 and y = x
h^-1 (x) = ¬{{x + 7}/2} - 26Example 10 Find the inverse of the function f(x) = 3/{x - 1}, {xER, x =/ 1}, by changing the subject of the formulaLet y = f(x)
y = 3/{x - 1}
y(x - 1) = 3
yx - y = 3
yx = 3 + y
x = {3 + y}/y
Therefore f^-1 (x) = {3 + x}/x - 27Example 11 The function f(x) is defined by f(x) = ¬{x - 2} {xER, x >= 2}. Find the function f^-1 (x) in a similar form stating its domainLook at textbook for the graphs
The range of the function is f(x) >= 0. Therefore the domain of the inverse function is x >= 0.
y = ¬{x - 2}
y^2 = x - 2
x = y^2 + 2
Therefore f^-1 (x) = x^2 + 2 - 28Example 12 If g(x) is defined as g(x) = 2x - 4 {xER, x >= 0} a) Calculate g^-1 (x) b) Sketch the graph of both functions on the same set of axes c) What is the connection between the graphs?a) g(x) = 2x - 4
Let y = g(x)
y = 2x - 4
y + 4 = 2x
{y + 4}/2 = x
f^-1 (x) = {x + 4}/2, {x >= -4}
b) Look at textbook for the graphs
c)The graphs of g^-1 (x) and g(x) are mirror images of each other in the mine y = x - 29Example 13 The function f(x) is defined by f(x) = x^2 - 3 {xER, x >= 0} a) Find f^-1 (x) in similar terms b) Sketch f^-1 (x) c) FInd values of x such that f(x) = f^-1 (x)a) Let y = f(x)
y = x^2 - 3
y + 3 = x^2
¬{y + 3} = x
f^-1 (x) = ¬{x + 3}, {xER, x >= -3}
b) Look at the textbook for the graph
c) f(x) = f^-1 (x)
f(x) = x
x^2 - 3 = x
x^2 - x - 3 = 0
So x = {1 + ¬{13}}/2 - 30The exponential and log functionsExponential functions occur naturally in real life scientists can model the number of elephants in a herd by using an exponential function
- 31Example 1 Sketch the graph of f(x) = 2^x for the domain xER. State the range of the functionLook at the textbook for the graph
The range of the function is f(x) > 0 - 32Example 2 Draw the graph of: a) y = e^x b) y = e^-xLook at the textbook for the graphs (answers)
- 33Example 3 Draw graphs of the exponential functions: a) y = e^2x b) y = 10e^-x c) y = 3 + 4e^({1}/{2}x)a)y = e^2x
= (e^x)^2
Look at the textbook for the graph
b) Look at the textbook for the graph
c) Look at the textbook for the graph - 34Example 4 The price of a used car can be represented by the formula, P = 16,000e^(t/10), where P is the price in £is and t is the age in years from new Calculate: a) the new price b) the value at 5 years old c) what the model suggests about the eventual value of the car Use this to sketch the graph of P and ta) Substitute t = 0 into P = 16,000e^(-0/10)
= 16,000 * 1
= £16,000
b) Substitute t = 5 into P = 16,000e^(-5/10)
=16,000e(-0.5)
= £9704.49
c) As t is infinate, e^(-t/10) = 0
Therefore P = 16,000 * 0 = 0
The eventual value is zero - 35Example 5 Slove the equations: a) e^x = 3 b) ln(x) = 4a) When e^x = 3
x = ln(3)
b) When ln(x) = 4
x = e^4 - 36Example 6 Sketch the graphs of: a) y = ln(3 - x) b) y = 3 + ln(2x)a) Look at the graph in the textbook
b) Look at the graph in the textbook - 37Example 7 Solve the equations: a) e^(2x + 3) = 4 b) 2ln(x) + 1 = 5a) e^(2x + 3) = 4
2x + 3 = ln(4)
2x = ln(4) - 3
x = {ln(4) - 3}/2
x = {ln(4)}/2 - 3/2
x = ln(2) - 3/2
b) 2ln(x) + 1 = 5
2ln(x) = 4
ln(x) = 2
x = e^2 - 38Example 8 (Part 1) The number of elephants in a herd can be represented by the equation: N = 150 - 80e^(-t/40) where N is the number of elephants in the herd and t is the time in years after the year 2003 Calculate: a) the number of elephants in the herd in 2003 b) the number of elephants in the herd in 2007N = 150 - 80e^(-t/40)
a) In the year 2003, N = 150 - 80e^(-t/40)
= 150 - 80*1
=70
b) In the year 2007, N = 150 - 80e^(-4/40)
=150 - 72.4
=78 (to the nearest elephant) - 39Example 8 (Part 2) The number of elephants in a herd can be represented by the equation: N = 150 - 80e^(-t/40), where N is the number of elephants in the herd and t is the time in years after the year 2003 Calculate: c) the year when the population will grow to above 100 d) the long-term population of the herd as predicted by the model Use all of the above information to sketch a graph of N against t for the modelc) If population is 100, then:
100 = 150 - 80e^(-t/40)
80e^(-t/40) = 50
e^(-t/40) = 50/80
-t/40 = ln(50/80)
t = -40ln(50/80)
t = 18.8 years
Therefore the population will be over 100 by the year 2022
d) As t = infinity, e^(-t/40) = 0 and therefore
N = 150 - (80*0)
The long term population as predicted by the model is 150
Look in the textbook for the graph - 40Numerical methodsThe branch of Mathematics called Numerical Analysis predates the invention of modern computers by many centuries. Many equations don't have exact solutions and iterative methods form successive approximations that converge to the exact solution
- 41Example 1 Show that the equation x^3 + 3x - 4 = 0 has a root between x = 2 and x = 3Look in the textbook for the graph
There is a root of the equation y = x^3 + 3x - 4 between x = 2 and x = 3 as there is a sign change - 42Example 2 Show that the values of y for points on the graph of y = 4 + 2x - x^3 change sign as the graph crosses the x-axisLook in the textbook for the graph
There is a change of sign as the graph crosses the x-axis - 43Example 3 Show that e^x + 2x - 3 = 0 has a root between x = 0.5 and x = 0.6Let f(x) = e^x + 2x - 3
f(0.5) = -0.351
f(0.6) = 0.022
There is a root between x = 0.5 and x= 0.6 as there is a change of sign - 44Example 4 a) Using the same axes, sketch the graphs of y = ln(x) and y = 1/x. Hence show that the equation ln(x) = 1/x has only one root b) Show this root lies in the interval 1.7 < x < 1.8a) Look in the textbook for the graph
The equation ln(x) = 1/x has only one root as the graphs have only one point of interception
b) ln(x) = 1/x
ln(x) - 1/x = 0
Let f(x) = ln(x) - 1/x
f(1.7) = -.0.0576
f(1.8) = 0.0322
There is a change of sign between f(1.7) and f(1.8), so the root of the equation ln(x) = 1/x lies inthe interval 1.7 < x < 1.8 - 45Example 5 a) Show that x^2 -4x + 1 can be written in the form x = 4 - (1/x) b) Use the itreration x(n+1) = 4 - (1/x(n)) to find, to 2d.p., a root of the equation x^2 -4x + 1 = 0. Start with x(0) = 3 c) Show graphically the first two iterations of this formulaa) x^2 -4x + 1 = 0
x^2 + 1 = 4x
x^2 = 4x - 1
x = 4 - (1/x)
b) x(1) = 4 - (1/x(0))
= 4 - 1/3
= 3.666 666 667
x(2) = 3.727 272 727
x(3) = 3.731 707 317
x(4) = 3.732 026 144
x(5) = 3.732 049 037
So a root is 3.73 to 2d.p.
c) Look at textbook for the graph - 46Example 6 (Part 1) a) Show that x^2 -5x -3 = 0 can be written in the form i) x = ¬{5x + 3} ii) x = {x^2 -3}/5a)i) x^2 -5x -3 = 0
x^2 - 3 = 5x
x^2 = 5x + 3
so x = ¬{5x + 3}
ii) x^2 -5x -3 = 0
x^2 -3 = 5x
{x^2 -3}/5 = x - 47Example 6 (Part 2) b) Show that the iteration formulae i) x(n+1) = ¬{5x(n) + 3} give different roots of the equation x^2 -5x -3 = 0. Start each iteration with x(0) = 5b)i) x(1) = ¬{5x(0) + 3}
=¬{5(5) + 3}
= 5.291 502 622
x(2) = 5.427 477 601
x(3) = 5.489 753
x(4) = 5.518 039 96
x(5) = 5.530 840 786
x(6) = 5.536 623 875
So - 48Example 6 (Part 3) b) Show that the iteration formulae ii) x(n+1) = {x(n)^2 - 3}/5 give different roots of the equation x^2 -5x -3 = 0. Start each iteration with x(0) = 5ii) x(1) = {x(n)^2 -3}/5
= {(5)^2 -3}/5
= 4.4
x(2) = 3.272
x(3) = 1.541 196 8
x(4) = -0.124 942 484 7
x(5) = -0.596 877 875 1
x(6) = -0.528 747 360 4
x(7) = -0.544 085 245 8
x(8) = -0.540 794 249 1
So a root is -0.5 to 1d.p.
Each iteration formula gives a sequence that converges to a different root of the equation - 49Example 7 (Part 1) a) Show that x^3 -3x^2 -2x +5 = 0 has a root in the interval 3 < x < 4 b) Use the iteration formula x(n+1) = ¬{{x(n)^3 -2x(n) +5}/3} to find an approximatation for the root of equation x^3 -3x^2 -2x +5 = 0. i) x(0) = 3a) Let f(x) = x^3 -3x^2 -2x +5
f(3) = -1
f(4) = 13
f(3) < 0 and f(4) > 0, so there is a change of sign. There is a root in the interval 3 < x < 4
b)i) x(1) = ¬{{x(0)^3 -2x(0) +5}/3}
= ¬{{(3)^3 -2(3) +5}/3}
= 2.943 920 289
x(2) = 2.865 084 947
x(3) = 2.756 113 603
x(4) = 2.609 192 643 etc
x(14) = 1.203 042 309
x(15) = 1.202 094 215
So a root is 1.2 to 1d.p. - 50Example 7 (Part 2) b) Use the iteration formula x(n+1) = ¬{{x(n)^3 -2x(n) +5}/3} to find an approximatation for the root of equation x^3 -3x^2 -2x +5 = 0. ii) x(0) = 4ii) x(1) = ¬{{x(0)^3 -2x(0) +5}/3}
x(1) = ¬{{(4)^3 -2(4) +5}/3}
x(1) = 4.509 249 753
x(2) = 5.405848 031
x(3) = 7.121 901 523
x(4) = 10.831 891 06
x(5) = 20.447 008 93
x(6) = 53.268 486 74
No root is found due to the fact each iteration gets further from a root, so the sequence is divergent
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- Descripción
- AS level Maths Fichas sobre Core 3 examples, creado por Benjamin Smith el 13/10/2017.
- Creado por
- Benjamin Smith
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