C3 June 2016 Q6

Notes

M1: Divides \(x^4 + x^3 - 3x^2 + 7x - 6\) by \(x^2 + x - 6\) to get a quadratic quotient and a linear or constant remainder. To award this look for a minimum of the following
\(\begin{array}{r} \mathbf{x^2 (+..x) + A}\phantom{{} + 7x - 6} \\ x^2 + x - 6\,\big)\overline{\,x^4 + x^3 - 3x^2 + 7x - 6} \\ \mathbf{x^4 + x^3 - 6x^2}\phantom{{} + 7x - 6} \\ \underline{\phantom{3x^2 + 7x - 6}} \\ \mathbf{(Cx) + D} \end{array}\)
If they divide by \((x + 3)\) first they must then divide their by result by \((x - 2)\) before they score this method mark. Look for a cubic quotient with a constant remainder followed by a quadratic quotient and a constant remainder
Note: FYI Dividing by \((x + 3)\) gives \(x^3 - 2x^2 + 3x - 2\) and \(\left(x^3 - 2x^2 + 3x - 2\right) \div (x - 2) = x^2 + 3\) with a remainder of 4.
Division by \((x - 2)\) first is possible but difficult.....please send to review any you feel deserves credit.

A1: Quotient = \(x^2 + 3\) and Remainder = \(4x + 12\)

M1: Factorises \(x^2 + x - 6\) and writes their expression in the appropriate form.
\(\left(\dfrac{x^4 + x^3 - 3x^2 + 7x - 6}{x^2 + x - 6}\right) \equiv \text{Their Quadratic Quotient} + \dfrac{\text{Their Linear Remainder}}{(x + 3)(x - 2)}\)
It is possible to do this part by partial fractions. To score M1 under this method the terms must be correct and it must be a full method to find both "numerators"

A1: \(x^2 + 3 + \dfrac{4}{(x - 2)}\) or \(A = 3, B = 4\) but don't penalise after a correct statement.

Alt (a) attempted by equating terms.

1st Mark   M1 Scored for multiplying by \((x^2 + x - 6)\) and cancelling/dividing to achieve
\(x^4 + x^3 - 3x^2 + 7x - 6 \equiv (x^2 + A)(x^2 + x - 6) + B(x \pm 3)\)
3rd Mark   M1 Scored for comparing two terms (or for substituting two values) AND solving simultaneously to get values of \(A\) and \(B\).
2nd Mark   A1 Either \(A = 3\) or \(B = 4\). One value may be correct by substitution of say \(x = -3\)
4th Mark   A1 Both \(A = 3\) and \(B = 4\)

Notes

M1: \(x^2 + A + \dfrac{B}{x - 2} \to 2x \pm \dfrac{B}{(x - 2)^2}\)
If they fail in part (a) to get a function in the form \(x^2 + A + \dfrac{B}{x - 2}\) allow candidates to pick up this method mark for differentiating a function of the form \(x^2 + Px + Q + \dfrac{Rx + S}{x \pm T}\) using the quotient rule oe.

A1ft: \(x^2 + A + \dfrac{B}{x - 2} \to 2x - \dfrac{B}{(x - 2)^2}\) oe. FT on their numerical \(A\), \(B\) for for \(x^2 + A + \dfrac{B}{x - 2}\) only

M1: Subs \(x = 3\) into their \(\mathrm{f}'(x)\) in an attempt to find a numerical gradient

M1: For the correct method of finding an equation of a normal. The gradient must be \(-\dfrac{1}{\text{their f}'(3)}\) and the point must be \((3, \mathrm{f}(3))\). Don't be overly concerned about how they found their f(3), ie accept \(x\)=3 y =.
Look for \(y - \mathrm{f}(3) = -\dfrac{1}{\mathrm{f}'(3)}(x - 3)\) or \((y - \mathrm{f}(3)) \times -\mathrm{f}'(3) = (x - 3)\)
If the form \(y = mx + c\) is used they must proceed as far as \(c =\)

A1: cso \(y - 16 = -\dfrac{1}{2}(x - 3)\) oe such as \(2y + x - 35 = 0\) but remember to isw after a correct answer.

Alt (b) is attempted by the quotient (or product rule) — 1st 3 marks

M1: Attempt to use the quotient rule \(\dfrac{vu' - uv'}{v^2}\) with \(u = x^4 + x^3 - 3x^2 + 7x - 6\) and \(v = x^2 + x - 6\) and achieves an expression of the form \(\mathrm{f}'(x) = \dfrac{\left(x^2 + x - 6\right)\left(..x^3........\right) - \left(x^4 + x^3 - 3x^2 + 7x - 6\right)(..x..)}{\left(x^2 + x - 6\right)^2}\).
Use a similar approach to the product rule with \(u = x^4 + x^3 - 3x^2 + 7x - 6\) and \(v = \left(x^2 + x - 6\right)^{-1}\)
Note that this can score full marks from a partially solved part (a) where \(\mathrm{f}(x) \equiv x^2 + 3 + \dfrac{4x + 12}{x^2 + x - 6}\)