Linear Inequalities: Eduqas GCSE Higher November 2017 Paper 1 Q10

Higher November 2017 Paper 1 Q10

10.

(a) Olly is trying to solve this inequality.\[10 - 2x \gt 3\]Here is Olly’s solution.

Is Olly’s solution correct?
Show clearly how you decide. [2]

(b)

(i) Sketch the graph of \(y = x^2 - 4\) for values of \(x\) between \(-3\) and 3 on the axes below. Mark any intercepts with the \(x\)-axis. [2]

(ii) Solve \(x^2 - 4 \lt 0\). [2]

Mark scheme (a)

Scheme	Marks	Comments
No (stated or implied) AND either a correct justification e.g. a comment such as ‘He should have reversed the inequality sign in step 3 because he divided by \(-2\)’ or ‘it should be \(x \lt \dfrac{-7}{-2}\quad x \lt 3.5\)’ or showing by substitution an example of a value of \(x \gt 3.5\) is not a solution of the original inequality or showing by substitution an example of a value of \(x \lt 3.5\) is a solution of the original inequality	E2	E1 for No and a partially correct justification. e.g. Stating that step 3 is incorrect. or stating that e.g. \(x = 4\) is not a solution (no subst seen) or stating the answer should be \(x \lt 3.5\) or stating that e.g. \(x = 2\) is not a solution (no subst seen)

Corrected by Westie's Workshop

The board's mark scheme printed “a value of of \(x \lt 3.5\)”. Changed to “a value of \(x \lt 3.5\)” because the word “of” was printed twice.

Original: see the board's mark scheme PDF (link above).

Possible misprint (note from Westie's Workshop)

One E1 line says “stating that e.g. \(x = 2\) is not a solution”. But \(10 - 2(2) = 6 \gt 3\), so \(x = 2\) is a solution of the original inequality, even though it is not in Olly’s answer \(x \gt 3.5\). The board probably means either of those points.

We have left the board's wording exactly as printed. Original: see the board's mark scheme PDF (link above).

Mark scheme (b)

Scheme	Marks	Comments
(i) Correct parabola through \((-2, 0)\) and \((2, 0)\).	B2	B1 for correct shape with intercepts relatively correct but roots not marked or for correct roots seen but shape of curve incorrect. Be generous with symmetry; ignore coordinates of vertex
(ii) \(-2 \lt x \lt 2\) or \(x \in (-2, 2)\)	B2	Accept \(-2 \lt x\) and \(x \lt 2\) or \(-2 \lt x,\ x \lt 2\) or the interval \((-2, 2)\) for 2 marks. B1 for each correct end or for \(-2 \lt x\) or \(x \lt 2\) or for ‘their \(-2\)’ \(\lt x \lt\) ‘their 2’, FT their intercepts from (b)(i) or for \(-2 \leqslant x \leqslant 2\) or for the correct region on the graph in (b)(i) identified as the solution set (including open circles at each \(x\)-intercept)
	(6)