Core 1 (C1)

    Mathematical Processes

      p1 - 3 Notation:

      • = equals
      • ≠ doesn't equal
      • ≡ identically equals (for all values of any terms in the equation)
      • ∴ therefore
      • ∵ because
      • ⇒ implies
      • ⇐ is implied by
      • ⇔ implies and is implied by
      • A condition is sufficient if it is true, but not for all values and a condition is necessary if it is true for all values of the statement
        - For example, to show that an integer is divisible by 5, it is sufficient if it ends in a zero (because integers ending in a 5 are also divisible by 5), and necessary that it ends in a 5 or 0
      • An expression is a phrase consisting of terms and operators (but no equals), such as 3×53 \times 5
      • A function takes an input to produce a single output. For example, f(5)f(5) takes an input (5 in this case) and outputs another number. What is does is unknown, and represented by ff

      Algebra

        a1 - 5, a8 - 9 Basics:

        • Be able to change the subject of a formula
        • Be able to graphically solve an equation
        • Be able to solve linear simultaneous equations with two unknowns (solve with subtraction or setting one equal to the other)
        • Be able to solve simultaneous equations where one equation is linear and one quadratic (solve by rearranging and then substituting the linear equation into the quadratic)
        • Find the points of intersection of two lines by solving the equations simultaneously
        • Be able to solve quadratic equations
          - By factorising:
          If a=1a = 1, find a pair of factors of cc which sum to the coefficient for b.
          If a>1a > 1, one method is to draw a 2×22 \times 2 table. Write the squared term (e.g. 3x23x^2) in the top left box and the constant term (e.g. 15) in the bottom right. Write down a list of all factors of (a×ca \times c) which sum to the second term of the quadratic. Take a pair and write them in the remaining two boxes and append an xx. If (a×ca \times c) is negative, choose a pair with one negative and one positive. Finally, find factors of the rows and columns. These make up the final factorised quadratic.
          By setting each bracket to equal zero and solving for xx, the two solutions can be found

          - With the quadratic formula:
          x=−b±b2−4ac2ax = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

          - By completing the square
          Subtract the third (constant term) from both sides. Take the second coefficient, half it and then square the answer. Add this value to both sides (with no xx). Now, the left hand side can be factorised and easily solved by square rooting both sides and solving for xx
          If a>1a > 1, it is easier to factorise out aa first. For example, for the equation 3x2−12x−7=03x^2 - 12x - 7 = 0:
          1. Move the constant term to the right-hand side: 3x2−12x=73x^2 - 12x = 7
          2. Factorise the left side, leaving blanks for the two constants (to be calculated later): 3(x2−4x+_)=7+_3(x^2 - 4x + \_) = 7 + \_
          3. Halve and square the second coefficient: =(4/2)2=4= (4/2)^2 = 4, 3(x2−4x+4)=7+_3(x^2 - 4x + 4) = 7 + \_
          4. Remember that the +4 is multiplied by the 3 outside the brackets, so add 4×34 \times 3 to the right side: 3(x2−4x+4)=7+123(x^2 - 4x + 4) = 7 + 12
          5. This can be factorised and simplified to 3(x−2)2=193(x - 2)^2 = 19

        a6 Discriminants:

        • The discriminant is calculated with the part of the quadratic formula under the square root, b2−4acb^2 - 4ac
        • The discriminant of a quadratic shows the number of real roots
          - If it is positive, there are 2 real roots and 2 xx-intercepts
          - If it = 0, there is one real root (repeated twice) and one xx-intercept
          - If it is negative, there are no real roots

        a7 Line of symmetry and turning point:

        • The xx-coordinate for the line of symmetry can be calculated with the formula x=−b÷2ax = -b \div 2a
        • The below examples use a completed square quadratic, (x+a)2+b(x + a)^2 + b
          - The turning point, which lies on the line of symmetry, is (−a,b)(-a, b). Therefore, the line of symmetry is x=−ax = -a from a completed square

        a11 - 12 Inequalities:

        • When multiplying or dividing both sides of an inequality by a negative number, the sign must be reversed (e.g. from a > / ≥ to a < / ≤ or vice-versa)
        • Quadratic inequalities are easier solved by first sketching the graph. Change the inequality symbol to an equals and solve the equation. Then re-add an inequality symbol based on the result of the sketch
          - For example, for the inequality x2−x−12>0x^2 - x - 12 > 0:
             - Factorise: (x−4)(x+3)>0(x - 4)(x + 3) > 0
             - Replace > with equals: (x−4)(x+3)=0(x - 4)(x + 3) = 0 ∴ x=4,x=−3x = 4, x = -3
             - Check the sketch to see where on the line y is greater than 0 (because the inequality is > 0)
             - The solution is therefore x<−3x < -3 or x>4x > 4

        a13 - 14 Manipulating surds:

        • a×b=ab\sqrt{a} \times \sqrt{b} = \sqrt{ab}
        • Surds can be simplified as follows (example):
          - 98\sqrt{98}:
          - Find a square factor of 98
          - 492=98\sqrt{49} \sqrt{2} = \sqrt{98}
          - Therefore, 72=987\sqrt{2} = \sqrt{98}
        • nn=n\sqrt{n}\sqrt{n} = n can be rearranged to get:
          n=nn\sqrt{n} = \frac{n}{\sqrt{n}} and 1n=nn\frac{1}{\sqrt{n}} = \frac{\sqrt{n}}{n}
        • This is used to rationalise a denominator, which is done by multiplying the numerator and denominator by bb if there is only a surd or a−ba - b if there is a rational number and a surd. (in this example, a is the rational number and b is the square root)
        • Remember that ab=(a)b\sqrt{a^b} = (\sqrt{a})^b

        a15 - 16 The laws of indices:

        • ya×yb=ya+by^a \times y^b = y^{a + b}
        • ya÷yb=ya−by^a \div y^b = y^{a - b}
        • (ya)b=ya×b( y^a )^b = y^{a \times b}
        • y0=1y^0 = 1
        • y−a=1yay^{-a} = \frac{1}{y^a}
        • y1a=y^{\frac{1}{a}} = a^ay\sqrt{y}
        • yab=(ya)1by^{\frac{a}{b}} = (y^a)^{\frac{1}{b}}

        Coordinate Geometry

          g1 - 14 Summary:

          • y−y1=m(x−x1)y - y_1 = m(x - x_1) where x1x_1 and y1y_1 are the coordinates for a point
          • Gradient = change in yy ÷ change in xx or y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}
          • The gradient of a perpendicular line is −1÷m-1 \div m where m is the gradient of the original line
          • The distance between two points can be calculated with Pythagoras' theorem
          • Find the intersection of two lines by setting them equal to each other or with simultaneous equations
          • x2+y2=r2x^2 + y^2 = r^2 is the equation of a circle with radius rr and centre (0,0)(0, 0)
          • (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2 is the equation of a circle with radius rr and centre (a,b)(a, b)
          • The angle in a semicircle is a right angle
          • The perpendicular from the centre of a circle to a chord bisects the chord (at a 90° angle)
          • The tangent to a circle at a point is perpendicular to the radius through that point

          Polynomials

            f1 Adding, subtracting and multiplying:

            • To add or subtract two polynomials, simply add/subtract the terms. For example:
              2x3+6x2+4x3+x2=6x3+7x22x^3 + 6x^2 + 4x^3 + x^2 = 6x^3 + 7x^2
            • When multiplying, each term in the first polynomial is multiplied by each term in the second polynomial. The result should then be simplified

            Dividing:

              Example:
              - f(x)=x3+x2−6xf(x) = x^3 + x^2 - 6x
              - g(x)=x−2g(x) = x - 2
              - Question: divide f(x)f(x) by g(x)g(x)
            • Step one:
              - The answer will be a quadratic in the form ax2+bx+cax^2 + bx + c, possibly with a remainder, dd
              - Therefore, (x−2)(ax2+bx+c)+d=x3+x2−6x(x - 2)(ax^2 + bx + c) + d = x^3 + x^2 - 6x
              - This can be expanded to ax3+bx2+cx−2ax2−2bx−2c+dax^3 + bx^2 + cx - 2ax^2 - 2bx - 2c + d
              - This simplifies to ax3−(b−2a)x2+(c−2b)x−2c+dax^3 - (b - 2a)x^2 + (c - 2b)x - 2c + d
            • Step two:
              - The order (power) of the first term in f(x)f(x) and the simplified expanded solution match, so they can be compared. They both have a coefficient of 1, so a=1a = 1
              - Similarly, (b−2a)(b - 2a) must equal 1, so we can deduce that b=3b = 3
              - (c−2b)=−6(c - 2b) = -6, so c=0c = 0
              - Since f(x)f(x) has no more terms, we stop here and conclude that d=0d = 0; there is no remainder. This could be confirmed beforehand by using the remainder theorem (see below)
            • Therefore, the answer is x2+3xx^2 + 3x, remainder 0

            f2 - f4 The factor theorem:

            • The factor theorem states that if f(a)=0f(a) = 0, then (x−a)(x - a) is a factor
            • By using different integer values for aa, the factors of a polynomial can be found
            • If not all factors can be found with this method, then division can be used to find all of them
              - Example: factorise x3–x2–3x+2x^3 – x^2 – 3x + 2
              - The only factor here that can be found with the factor theorem is (x−2)(x - 2)
              - Step one:
                 - Divide x3–x2–3x+2x^3 – x^2 – 3x + 2 by (x−2)(x - 2)
                 - This is x2+x–1x^2 + x – 1, with no remainder
              - Step two:
                 - Therefore, the factorised polynomial is (x−2)(x2+x−1)(x - 2)(x^2 + x - 1)
                 - This quadratic does not factorise, so this is the final answer
            • Only factors of the constant term of the polynomial might work as a value for aa
              - For example, with x3−x2−3x+2x^3 - x^2 - 3x + 2, the constant term is +2, so the values to try are 1, -1, 2 and -2

            f5 The remainder theorem:

            • When a polynomial, e.g. f(a)f(a), is divided by (x−a)(x - a), the remainder is always f(a)f(a)

            f6 Sketching:

            • A turning point is where a graph changes from decreasing to increasing, or increasing to decreasing (changing direction)
            • The graph of a polynomial of order n has a maximum of n−1n - 1 turning points
              - The order is the maximum power (e.g. the order of a cubic equation is 3, quadratic is 2, linear is 1)
            • The dominant term is the largest - e.g. the x3x^3 term in a cubic and x2x^2 term in a quadratic
            • If a polynomial has a positive dominant term, the yy value will approach infinity as xx is increased (the right 'tail' points upwards)
            • If a polynomial has a negative dominant term, the opposite happens - yy will approach negative infinity (the right 'tail' points downwards)
            • If the order is even, both 'tails' will point in the same direction. They will be in opposite y directions if odd
            • To find intersections with the xx axis, factorise and find the roots - for example, the roots and xx intersections of (x+1)(x−1)(x−3)(x + 1)(x - 1)(x - 3) are -1, 1 and 3
            • The constant term is the yy intercept

            f6 - 9 Binomial expansion:

            • Pascal's triangle is a triangle of numbers, starting with 1, where each number is equal to the sum of the two above it. The single 1 at the top is not considered a row. So the 'first row' is {1,1}\{1, 1\} and the 'second' is {1,2,1}\{1, 2, 1\}
            • This can be used to expand polynomials. For example: expand (x+3)4(x + 3)^4:
              - The fourth row of Pascal's triangle is {1,4,6,4,1}\{1, 4, 6, 4, 1\}
              - Therefore, the expansion is (1×x4×30)+(4×x3×31)+(6×x2×32)+(4×x1×33)+(1×x0×34)(1 × x^4 × 3^0) + (4 × x^3 × 3^1) + (6 × x^2 × 3^2) + (4 × x^1 × 3^3) + (1 × x^0 × 3^4)
              - This simplifies to x4+12x3+54x2+108x+81x^4 + 12x^3 + 54x^2 + 108x + 81
            • In questions like expand (2x+3y)3(2x + 3y)^3, remember that the first term of the expansion would be 1x(2x)3x3y01 x (2x)^3 x 3y^0, simplifying to 8x38x^3, not 2x32x^3 (the two must also be cubed as well as the xx)
            • Alternatively, use nCr^nC_r [also written (nr)\begin{pmatrix}n \\ r\end{pmatrix}]:
              - nCr=n!r!(n−r)!^nC_r = \frac{n!}{r!(n - r)!} where
              - nn is the order of the polynomial
              - rr is the term you want to find, counting from 0 - so to find the fourth term, use 3, and to find the 25th term, use 24
              - Multiply this by brb^r to get the final coefficient (where you are expanding (a+b)n(a + b)^n
            • The sum of all binomial coefficients is equal to 2n2^n, where nn is the polynomial order. In the above example for (x+3)4(x + 3)^4, the sum of binomial coefficients is 1+4+6+4+11 + 4 + 6 + 4 + 1 (row from Pascal's triangle). This can be quickly calculated with 2n2^n; 24=162^4 = 16