Pure Mathematics P1

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1 Pure Mathematics P1 Rules of Indices x m * x n = x m+n eg. 2 3 * 2 2 = 2*2*2*2*2 = 2 5 x m / x n = x m-n eg. 2 3 / 2 2 = 2*2*2 = 2 1 = 2 2*2 (x m ) n =x mn eg. (2 3 ) 2 = (2*2*2)*(2*2*2) = 2 6 x 0 = 1, x 0 eg. 2 2 = 1 2 2 = 2 0 2 2 2 2 x -m = 1 eg. 2-3 = 2 1 / 2 4 2/16 = 1/8 x m x 1/n = n x eg. x 1/2 * x 1/2 = x 1 Therefore x 1 = x 1/2 x m/n = n x m eg. x 2/3 * x 2/3 * x 2/3 = x 6/3 = x 2 Therefore 3 x 2 = x 2/3 Special Numbers i = -1 e = 2.7... Φ (phi) = Golden Ratio sin 45 = 1/ 2 Types of Numbers Check pg134 for notation Natural numbers 1, 2, 3, 4... Integers 0, ±1, ±2, ±3... Rational Numbers Q comes from quotient, another word for division Can be written in the form p / q where p and q Є, and q 0 Real Numbers All the numbers that exist on the number line Rules of Surds a b = ab a = a / b b m a + n a = (m=n) a Note: a ± b a±b In the formula for solving quadratics: b 2 4ac is called the discriminant Quadratics If the discriminant b 2 4ac < 0 there are no real roots If the discriminant b 2 4ac > 0 there are two real roots If b 2 4ac = 0 then there is one repeated root. Solving Simultaneous Equations Elimination Method Make coefficients of either x or y the same.

Signs the same subtract, signs different add. Remove one of the unknowns by adding or subtracting so you are left with only one unknown. Substitute in the value you have found to find the other unknown. Substitution Method When the coefficient of a constant is 1. Substitute in the value of one constant as its comparison to the other ie if x = y+1 then substitute any x for y+1 so you only have one unknown. Solve Alternative Method Shuffle the equation so that the left hand side of both equations is equal and contains one of the unknowns. Then take the right hand side of each and write as an equation. Solve this equation to find one of the unknowns then substitute in that value to find the other. Elimination Method with Products Instead of adding or subtracting as with the previous elimination mehtod, divide or multiply to find x and y. Solving Non-Linear Simultaneous Equations Always substitute the linear into the non-linear. Remember to take out common factors. Inequalities Linear Solve like a linear equation, except flip the sign around when multiplying or dividing by negative numbers. Always move smallest value. Quadratic As the inequality is not necessarily equal to zero, we cannot assume one of the brackets, when factorised, will be 0. We must transfer the inequality to an equation and find the roots. These are our Critical Values. Sketch the graph of y = the equation and mark on the roots. Depending on the inequality sign, find whether the range of values lies inside or outside this amount. You may have to satisfy two inequalities. Sketching Graphs Quadratics x intercepts are roots. Y intercept is constant. Lowest point can be found be completing the square (it is the number after the squared bracket, as the lowest the bracket can equal is 0) Cubics When x coefficient > 0 the graph goes from bottom left to top right. When x coefficient < 0 the graph goes from top left to bottom right. If the factorised form contains a (x+n) 2 then the graph just touches the x axis at -n. You will not be expected to find maxima and minima for curves other than quadratics. Will only be asked to factorise if there is no constant term, as everything will be divisible by x, and you will then just need to solve a quadratic. Sketching Graphs To sketch y = x 3 : Put y = 0 and solve for x. As the curve passes through the x axis at only one point, find its shape by plotting a 2

3 few points. Notice that as x increases, y increases rapidly. The curve is 'flat' at (0,0). This point is called the point of inflexion. The gradient is positive just before (0,0) and just after (0,0) Notice that the shape of the curve is the same as y = (x + 1) 3 y = -x 3 is a reflection in the x axis of y = x 3 y = (x + 1) 3 is a translation of y = x 3, moved -1 along the x axis. The point of inflexion is (- 1,0) y = (3 - x) 3 can be rewritten as y = - (x - 3) 3 so the curve will have the same shape as y = - x 3. It is a translation of +3 along the x axis, so the point of inflexion is at (3,0) To sketch y = k / x where k is a constant: The curve does not cross the axes The curve tends towards the x axis when x is large and positive or large and negative. The x axis is a horizontal asymptote. The curve tends towards the y axis when y is large and positive or large and negative. The y axis is a vertical asymptote. You can draw a dashed line to indicate an asymptote There are 2 categories of y = k / x graphs: Type 1 Type 2 k > 0 k < 0 1 1 0.7 0. 0.2 0.7 0. 0.2 0-0.25-0.5-0.75-1 -4-3 -2-1 0 1 2 3 4 0-0.25-0.5-0.75-1 -4-3 -2-1 0 1 2 3 4 Reflection in y = -x Reflection in y = x Sketch the two graphs on the same axes. Solving Equations and Finding Points of Intersection Often the equations are given explicitly. ie y =... If this is true, set the two functions equal to each other. Expand brackets and solve to find x values. Sub results into simplest of two equations to find y values. Write out the coordinates

4 Be careful not to divide by an unknown term eg x - 1, unless you are certain x - 1 cannot equal 0. Compared with a graph y = f(x) Transformations y = f(x) + a y = f(x + a) y = af(x) y = f(ax) is a translation of a units in the positive y-axis direction is a translation of a units in the negative x-axis direction is a stretch away from the x axis in the direction of the y axis by a scale factor of a is a stetch away from the y axis in the direction of the x axis by a scale factor of 1 / a Special Cases y = -f(x) y = f(-x) is a reflection in the x axis is a reflection in the y axis If a is in the bracket, it affects the x axis in the opposite way to expected. If a is outside the bracket, it affects the y axis in the way you would expect it to. Coordinate Geometry in the (x, y) plane (Straight line graphs) In the general form y = mx + c; m is the gradient, c is the y intercept + easy to see gradient and intercept - may be awkward numbers In the general form ax + by + c = 0; where a,b,c Є (where a, b, c are integers). It is often useful to rearrange into the form y = mx + c + all numbers are integers - harder to see gradient and intercept 1. Gradient You can work out the gradient m of the line by joining the point with coordinates (x 1,y 1 ) to the point with coordinates (x 2,y 2 ) by using the formula m = y 2 - y 1 x 2 - x 1 It doesn't matter which point is taken as (x 1,y 1 ) Don't use this formula if you are just trying to figure out the equation of the line. 2. Equation of a Line with Gradient m You can find the equation of a line with gradient m that passes through the point (x 1,y 1 ) by using the formula y - y 1 = m(x - x 1 ) 3. Equation of a Line when given two points

5 You can find the equation of a line that passes through the points with coordinates (x 1,y 1 ) and (x 2,y 2 ) by using the formula 4. Parallel and Perpendicular Lines Parallel lines have the same gradient but different y intercepts If a line has a gradient m, a line perpendicular to it would have a gradient of -1 / m The gradient of the two lines has a product of -1 y - y 1 = x - x 1 y 2 - y 1 x 2 - x 1 3, 5, 7, 9... is a sequence 3 + 5 + 7 +9... is a series Sequences and Series Each number in a sequence or series is called a TERM. The GENERAL TERM is also known as the n th term or the position to term rule This can also be written as U n = Recurrence Relations These rules define a sequence by relating each term to the previous term (or terms) eg. The Fibonacci Sequence U n =U n-1 + U n-2 Types Of Sequence 3, 5, 7, 9 Increasing Sequence --> 9, 7, 5, 3, 1, -1 Decreasing Sequence --> - 1/2, 1/4, 1/8 Converging Sequence -->Number (in this case 0) 1, -1, 1, -1, 1 Oscillating 1, 2, 3, 4, 3,2,1, 2 Periodic Arithmetic Progressions (APs) These are sequences where successive terms increase or decrease by a constant amount (known as the common difference) d is the common difference a is the first term Second term = a + d Third term = a + 2d Nth term = a + (n-1)d Sum of Arithmetic Progressions (APs) S n = a + (a+d) + (a+2d) +... + (a+(n-1)d) S n = f (a, d, n) S n = an + 1d + 2d +... + (n-1)d S n =an + n * (n 2 - n)*d/2 S n = n / 2 ( 2a + (n- 1) d) S n = n / 2 (a + L) a = first term L = last term S n - S n-1 = U n

6 n Σ 3r + 2 r=1 The Sigma Notation means sub in r=1 into 3r+2 then r=2, r=3,...r=n Find the sum of these terms In general and AP will be written in the form n Σ pr + q By First r=1 Principles p, q Є Differentiation AB and AC are chords. The gradient at A is closer to the gradient of AC than AB, since the chord AC is shorter. Every time one makes the chord shorter, the gradient of the chord gets closer and closer to the gradient of the curve at A. Eventually, when the chord becomes so short that it is a tangent, the gradient of the graph will equal the gradient of this tangent. Consider the point P(x, y) on y = x 2. It creates a chord with point Q(x + δx, y + δy) where δx means a small change in x Gradient PQ = δy / δx y= x 2 at Q: y + δy = (x + δx) 2 y + δy = x 2 + 2xδx + δx 2 δy = 2xδx + δx 2 δy / δx = 2x + δx As δx tends towards 0, δy / δx = 2x lim (δy) dx --> 0 (δx) = 2x dy dx = 2x We say "dy by dx" and it means the gradient function. If y = ax n dy dx = anx n-1 When the function is a sum, each term can be differentiated separately.

7 Always expand brackets first. Differentiation if y=f(x) ± g(x) then dy/dx = f'(x) ± g'(x) Second Order Derivative d 2 y/dx 2 = f''(x) "dee two why by de ex squared" When you differentiate an already differentiated expression. Rate of Change of a Function x is independent variable y is dependent variable. These variables often stand for quantities You can substitute: t for time V for volume P for population A for area r for radius s for displacement h for height v for velocity σ for temperature etc. Rate of change of velocity with respect to time is acceleration Rate of change of displacement with respect to time is velocity dv/dt = acceleration ds/dt = velocity Finding the Equation of a Tangent and its Normal The tangent at point (a, f(a)) has gradient f'(a) Can find its equation using: y - y 1 = m (x - x 1 ) where m is f'(a) Therefore y - f(a) = f'(a) (x - a) The normal to the curve at point A is defined as being the straight line through A which is perpendicular to the tangent at A. Gradient of normal = -1/f'(a) Equation of normal = y - f(a) = -1/f'(a) (x - a) Integration

8 Translations of a graph have the same gradient function, so when you reverse differentiation you end up with y = f(x) + c where c can be any number C Є Integration allows us to find areas under curves, however this is not until C2 To note down "the integral of y" we write y dx Which means "the integral of y with respect to x" In general ax n dx = ax n+1 + c n + 1 As with differentiation, always get expression as a collection of ax n terms. Given dy/dx and the coordinates of a point on the curve, we can find the equation of the curve using integration and substitution.