A-LevelMathematicsAlgebra9709/22 · Oct/Nov 2022 · Q25 Marks
The solutions of the equation |4x − 1| = |x + 3| are x = p and x = q, where p < q. Find the exact values of p and q, and hence determine the exact value of |p – 2| – |q – 1|.
- (a)The solutions of the equation |4x − 1| = |x + 3| are x = p and x = q, where p < q. Find the exact values of p and q, and hence determine the exact value of |p – 2| – |q – 1|.[5]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.
- [B1] Solve 4x − 1 = x + 3 to obtain x = 4/3
- [M1] Attempt solution of linear equation where signs of 4x and x are different
- [A1] Obtain final value x = −2/5
- [M1] Substitute numerical values and apply modulus signs correctly to obtain |4/3 − 2| − |−2/5 − 1| or equivalent, retaining exactness and with no subsequent squaring Allow their p and q, p < q.
- [A1] Obtain 31/15 or exact equivalent.
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