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OCR A Level Math - Pure
Factor Theorem - ANS-A polynomial f(x) has a factor (x - a) if and only if f(a) = 0 It has a factor (bx-a) if f(a/b) = 0 Sketching polynomials - ANS-Shape: Reflect in x-axis if there is a negative x Roots: - Crosses x-axis at any root e.g. (x+5) would mean crosses at negative 5 -However, if negative x is in the bracket e.g. (5-x) would mean crosses at positive 5 -Also a repeated root means the curve just touches at that point Sketching graphical inequality - ANS-Shade where the inequality is not satisfied and use the origin to determine this Transformation of curves - ANS-y = f(x)+a: Translation by 'a' units parallel to the y-axis y = f(x-a): Translation by 'a' units parallel to the x-axis y = af(x): Stretch parallel to the y-axis by a scale factor of 'a,' (1,3) goes to (1,3a) y = f(ax): Stretch parallel to the x-axis by a scale factor of '1/a,' (1,3) goes to (1/a,3) y=-f(x): Reflect f(x) in x-axis y=f(-x): Reflect f(x) in y-axis How to determine a stationary point - ANS-dy/dx = 0 How to determine what a stationary point is - ANS-Calculate d2y/dx If negative, local maximum If positive, local minimum If 0, can be max,min or point of inflection What is a point of inflection? - ANS-A point where the graph of f changes from concave upward to concave downward or vice versa. So a point when a gradient changes from increasing to decreasing or vice versa How to determine if there is a point of inflection - ANS-If d2y/dx2 = 0, there could be If dy/dx ≠ 0, it is a non horizontal point of inflection If dy/dx = 0, test to see if there is a sign change of dy/dx either side of the point, if there is not, it is a horizontal point of inflection.
Optimisation - ANS-If asked to calculate minimum/maximum value of something Differentiate and set to equal 0, to find the value of the letter the something is in terms of Check if it is max/min by using the 2nd derivative and plugging in the letter value What does dy/dx show? - ANS-Rate of change of y wrt x (gradient) What does d2y/dx2 show? - ANS-Rate of change of gradient wrt x decreasing function - ANS-Graph that falls from left to right dy/dx < 0 increasing function - ANS-from left to right, the graph goes up. dy/dx > 0 How to find other value of trigonometric function - ANS-sinx = sin(180-x) cosx = cos(360-x) = cos(-x) tanx = tan(x+180) sin related to cos - ANS-sinx = cos(90-x) cosx = sin(90-x) 2 trig identities - ANS-tanx = sinx/cosx cos2x + sin2x = 1 Concentric circles - ANS-circles that lie in the same plane and have the same center Equation of a circle - ANS-(x-a)²+(y-b)²=r² Discriminant - ANS-b^2-4ac What does discriminant show - ANS-b^2-4ac <0, no real solutions b^2-4ac =0, 1 repeated solutions b^2-4ac >0, 2 distinct solutions Graph of 1/x - ANS- Graph of x^3 - ANS- Graph of a^x or e^x - ANS-
Prove by contradiction that √2, pi etc. is an irrational number - ANS-Assume it is rational so √2 =a/b, in its simplest form Square both sides leading to 2b^2 = a^2, as a^2 is even, a is even. So let a = 2k Now show b^2 = 2k^2, so b is even But as they are both even they cannot be in the simplest form So you have proven it Criticising proofs caution - ANS--Careful with logs as you cannot take logs of a negative -Careful when dividing by a variable/trig function as the trig function could = Sketching modulus functions - ANS-Any part of the graph that would go under the x- axis, is reflected in the x-axis Transforming modulus function - ANS-f(x) = I...I, so if asked to translate into another mod function you have to use f(x-a) as f(x) + a would not be in the modulus signs. Solving modulus functions - ANS-Solve for when positive and for when negative. Arithmetic Sequences - ANS-Can be found by adding the same number to the previous term Arithmetic Sequence nth term Formula - ANS-Un=a+(n-1)d a= first term d= common difference Geometric Sequence - ANS-a sequence in which each term is found by multiplying the previous term by the same number Geometric Sequence nth term Formula - ANS-an=a(r)^n- r=common ratio Convergent geometric series - ANS-A geometric series where and the sum gets closer and closer to some real number, has to have |r|< Sigma Notation - ANS- Partial Fractions - ANS-Multiply each of the new fractions, by the original denominator Compare coefficients Partial Fraction Rule for repeated brackets - ANS-A/(x+p)^2 = B/(x+p) + C/(x+p)^ Partial Fraction rule for power of x - ANS--If the power of x is the same on top and bottom, and the coefficients of the largest x degree is, are different -Take away the coefficient of the x in terms of the denominator, so that the coefficients of the greatest x power are same top and bottom.
e.g. (mx^2)/(x^2 - a) - m(x^2 - a)/(x^2 - a) = ma/(x^2 - a) One to one vs Many to one function - ANS-One to one: Every y value only has 1 possible x value Many to one: Y values can have multiple x values How to tell if a function has an inverse - ANS-If it is one to one it has an inverse, if it is many to one, it doesnt How to calculate inverse - ANS-Swap x and y Make y the subject of the formula and that is the inverse How to prove a function is one to one? - ANS-Show gradient is always positive/negative. What does inverse function mean on a graph - ANS-Reflected in y=x How to rearrange into the form (1+x)^n for binomial expansion - ANS-(1-x/1+x)^0.5 = (1- x)^0.5 * (1+x)^-0. (9+x)^0.5 = [9(1+x/9)]^0.5 = 3(1+x/9)^0. Restriction to make a binomial expansion convergent - ANS-For (1+x)^n IxI < 1, IF ASKED FOR UR VALUE MAKE SURE U KEEP THE MOD SIGNS Radians to degrees - ANS-2π= 360° Transformations for trigonometric graphs - ANS-y = A sin/cos k(x+Φ) + d |A| = Amplitude 2π/k = Period d = Central Value Φ = Phase sinx + cosx identities - ANS-asinx + bcosx = Rsin(x+α) asinx - bcosx = Rsin(x-α) How to do sinx + cosx questions - ANS-Use the identities Then use formula book identities for the brackets and equate the coefficients, to get Rcosα=a and Rsinα=b. (Careful when using the formula book identities and make sure you get the negatives right)
arcsin graph - ANS- arccos graph - ANS- arctan graph - ANS- Arc Length Formula - ANS-rθ (in radians) Sector Area Formula - ANS-1/2 r^2θ (in radians) Segment Area Formula - ANS-area of sector - area of triangle 1/2 r^2θ - 1/2 r^2 sinθ (in radians) Derivative of sinkx - ANS-kcoskx Derivative of coskx - ANS--ksinkx Derivative of e^f(x) - ANS-f'(x)e^f(x) Derivative of a^x - ANS-a^x ln(a) Derivative of loga(x) - ANS-1/xlna Chain rule - ANS-dy/dx = dy/du * du/dx Product Rule - ANS-dy/dx = uv' + vu' Implicit Differentiation - ANS-1. Differentiate each term with respect to x. Numbers go to 0
- Use power rules on y terms but multiply by dy/dx
- If you have an x and y together, use product rule Integral of (ax+b)^n - ANS-1/a(n+1) * (ax+b)^n+1 +c Integral of 1/(ax+b) - ANS-1/a * ln(ax+b) + c How to integrate x/(x+a) - ANS-Make it into (x+a-a)/(x+a) = 1 - a/(x+a) integral of cos(kx) - ANS-1/k sin(kx) + c integral of sin(kx) - ANS--1/k coskx + c integral of sec^2(kx) - ANS-(1/k)tan(kx) + C integral of secx - ANS-ln|secx+tanx|+C
(Multiply by secx+tanx/secx+tanx) Integration by substitution - ANS-Use u to find dx in terms of du Integrate the term wrt u Sub the x term for u back in Integration by substituting trig values - ANS-Sometimes you are not given a trig substitution and have to think of your own When would you substitute asinu/acosu for an integral - ANS-∫√(a^2 - x^2) dx As sin^2 + cos^2 = 1 So you know sin or cos = √(1 - sin^2 or cos^2 ) When would you substitute atanu for an integral - ANS-∫√(a^2 + x^2) dx As tan^2 + 1 = sec^ So you know sec = √(1 + tan^2) Rule for subbing in trig for an integral - ANS-Look at trig identities and see if they will simplify the problem Integration by parts - ANS-∫ u dv = uv - ∫ v du (given) Integration by parts order of priority for choosing u - ANS-LATE Logs,Algebra,Trig,Exponential Is the integral a product - ANS-If a product of trig functions, use a trig identity to rewrite the expression If just a product, use integration by parts Is the integral a fraction - ANS-Is the top order higher than the bottom, use algebraic division then partial fractions if necessary Is the top order lower than the bottom, use partial fractions. Area enclosed by 2 curves, y=g(x), y=f(x) - ANS-Definite integral with 2nd intersection on top, 1st intersection on the bottom Integrate f(x)-g(x) Area bounded by y-axis - ANS-Write function with x as subject Integrate against dy with y limits
General solution of differential equations - ANS-Just y and x. Differential Equation Tips - ANS-Remember the +c with the integration, and if a constant has been taken out before in the integral, remember the +c is also multiplied by this. If there is proportionality, integrate with the k and find the c with the k first. Then after this, find the value of k. What should you remember when making a differential equation model - ANS- Conditions e.g. weather Numerical Methods of Solving Equations - ANS-Trial+Improvement Iteration Calculator Trial and Improvement - ANS-Use table function in calculator and see where it is close to the answer Keep making the step smaller until you get a 1dp answer Sign Change Method - ANS-Type of trial and improvement used to locate roots of f(x)= If you notice a change in sign of y-value then this indicates a root between the values of x as the root is where y= Sign Change Method Failures - ANS-The curve could touch the x-axis and there would be a root but no change of sign as both sides will be positive or both negative The sign change isn't noticed between points where there are multiple roots, for example the curve could dip under the x-axis and come back up, so have 2 roots but the change of sign may not be noticed as the points are either side of the dip. Simple iterative methods - ANS--Rearrange equation so x is the subject, x=g(x) -Make a guess and plug into g(x) then keep plugging answers back into g(x) -Should converge to a root Xn+1 = g(Xn) converges to a root at x=a, if g'(a)<1, and if X1 is close to a. Simple iterative methods, rearrangement failure - ANS-If g'(a)>1, it may work but not definitely. Therefore, if it doesn't work use another rearrangement to make x the subject. Simple iterative methods, staircase + failure - ANS-The staircase diagram shows how the initial guess of X0 tends towards the answer which is the intersection point of y=x and y=g(x). The diagram always starts by hitting y=g(x), then it builds a staircase.
Failure: y=g(x) could be a shape where no guess will ever tend towards the solution, e.g. a convex curve, and instead it will either go to another root or to infinity. Therefore, you have to use another rearrangement to make x the subject. Simple iterative methods, cobweb + failure - ANS-The cobweb diagram works the same as the staircase, but forms a cobweb instead. Failure: y=g(x) could be a shape and the guess could make a cobweb that comes exactly back on itself and forms an endless rectangle. This could be fixed with a different rearrangement or guess. Newton-Raphson Method - ANS-An iterative procedure for solving f(x)=0. Starts with a guess for a root and using the formula from the formula book, gets an answer closer and closer to the root and converges at the root. Other roots can be found with different guesses. Newton-Raphson Method Diagram - ANS-Starts at X0 (guess), straight up to f(x). A tangent at f(x) crosses X1 where the next vertical line is drawn for the next tangent. This tends towards the root Newton-Raphson Method Failures - ANS-If the guess is made at a stationary point, the tangent will be a flat line so will not converge to any root. If the guess makes the denominator in the equation = 0, the fraction will be undefined so X1 will be undefined as well so there will be no converging to a root. Consideration with iteration stuff - ANS-Your value might have to be an integer for example if it is applied Cartesian equation - ANS-A single equation involving just x and y Equation of a straight line - ANS-y - y1 = m(x - x1) y = mx+c Is 0 an integer - ANS-yes, but its neither +ve or -ve Prove ABCD is a parallelogram - ANS-Prove AB and DC are the same vector Prove ABCD is a rhombus - ANS-Prove AB and DC are the same vector Then show 2 adjacent (next to each other) side have the same magnitude. Equation of a circle with position vector p - ANS-p-(centre) = radius
