Simultaneous equations, inequalities and polynomials – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (5 marks)

   One equation is linear and one is not, so make \(y\) the subject of the linear equation and substitute.

   \(y=x-2\) (1). \(x(x-2)=15\) (1), \(x^2-2x-15=0\) (1), \((x-5)(x+3)=0\) (1). Solutions: \((5,3)\) and \((-3,-5)\) (1).

   For each \(x\) use \(y=x-2\): \(x=5\) gives \(y=3\) and \(x=-3\) gives \(y=-5\). Check: \(5\times3=15\) ✓ and \((-3)(-5)=15\) ✓.

   Pairing the values wrongly, for example \((5,-5)\), loses the last mark.

2. Answer to Question 2 (4 marks)

   Factorise the quadratic and use a sketch: a U-shaped graph is positive outside its roots. Then solve the linear inequality and combine the two.

   \((2x+1)(x-4)\ge0\) (1), so \(x\le-\tfrac12\) or \(x\ge4\) (1). \(3x-6\lt 2x+5\) gives \(x\lt 11\) (1). Both: \(x\le-\tfrac12\) or \(4\le x\lt 11\) (1).

   On a number line, the overlap is everything up to \(-\tfrac12\) together with the interval from 4 up to (but not including) 11.

   The value \(x=11\) is excluded because the linear inequality is strict.

3. Answer to Question 3 (4 marks)

   1. By the factor theorem, \((x-2)\) is a factor if \(f(2)=0\). \(f(2)=8-16+2+6=0\) (1) So \((x-2)\) is a factor.
   2. Divide \(f(x)\) by \((x-2)\), then factorise the quadratic quotient. \(f(x)=(x-2)(x^2-2x-3)\) (2) \(=(x-2)(x-3)(x+1)\) (1) So the roots of \(f(x)=0\) are \(-1\), 2 and 3. Check: \(f(3)=27-36+3+6=0\) ✓ and \(f(-1)=-1-4-1+6=0\) ✓.

   The quotient can also be found by comparing coefficients: \((x-2)(x^2+bx+c)\) gives \(b-2=-4\) and \(-2c=6\), so \(b=-2\) and \(c=-3\).

4. Answer to Question 4 (3 marks)

   By the remainder theorem, the remainder on dividing by \((x-2)\) is \(p(2)\), so set \(p(2)=7\).

   \(p(2)=16+2a-5=7\) (2), so \(2a=-4\) and \(a=-2\) (1).

   Check: \(p(2)=2\times8-2\times2-5=7\) ✓. So \(a=-2\).

   The remainder theorem avoids long division: substituting \(x=2\) gives the remainder directly.

   A common error is substituting \(x=-2\): the divisor \((x-2)\) is zero when \(x=2\). Setting \(p(2)=0\) instead of 7 confuses the remainder theorem with the factor theorem and gives the wrong value \(a=-5.5\).

5. Answer to Question 5 (4 marks)

   1. The length is \(w+3\), and area \(=\) width \(\times\) length. \(w(w+3)=54\) so \(w^2+3w-54=0\) (1) This is the required equation.
   2. Factorise and reject the negative root, since a width cannot be negative. \((w+9)(w-6)=0\) (1), width 6 m and length 9 m (1) Check: \(6\times9=54\) ✓.
   3. The perimeter is \(2(6+9)=30\) m; multiply by the cost per metre. Perimeter \(=30\) m, cost \(=30\times4.5=\) £135 (1) Edging the lawn costs £135.

Total : 20 marks