EDEXCEL AS LEVEL FURTHER MATHEMATICS PAPER 22 FURTHER PURE MATHEMATICS 2 NEW UPDATE WITH ALL COMPREHENSIVE REAL EXAM QUESTIONS AND A DETAILED BREAKDOWN OF ALL CORRECT VERIFIED ANSWERS EXAM PLUS CERTIFIED RATIONALES | COMPLETE TEST SOLUTION | PAS

EDEXCEL AS LEVEL FURTHER MATHEMATICS
PAPER 22 FURTHER PURE MATHEMATICS 2 NEW
2026-2027 UPDATE WITH ALL COMPREHENSIVE
REAL EXAM QUESTIONS AND A DETAILED
BREAKDOWN OF ALL CORRECT VERIFIED
ANSWERS EXAM PLUS CERTIFIED RATIONALES |
COMPLETE TEST SOLUTION | PASSED & REWARDED
WITH DISTINCTION FOR ORIGINALLY GRADE A+
BRAND NEW!!!




This paper covers the full official syllabus for
Further Pure Mathematics 2 (8FM0/22), including
inequalities, series, further complex numbers, first-
and second-order differential equations, Maclaurin
and Taylor series, and polar coordinates.




Section A: Inequalities (Questions 1–12)
1. Solve the inequality x+1x−3<2x−3x+1<2.
x<3x<3 or x>7x>7

,Multiply both sides by (x−3)2>0(x−3)2>0 to
preserve the inequality direction. This
gives (x+1)(x−3)<2(x−3)2(x+1)(x−3)<2(x−3)2.
Expand and
simplify: x2−2x−3<2x2−12x+18x2−2x−3<2x2−12x
+18, leading
to 0<x2−10x+21=(x−3)(x−7)0<x2−10x+21=(x−3)(
x−7). The quadratic is positive
when x<3x<3 or x>7x>7. Note that x=3x=3 is
excluded as it makes the original denominator
zero.
2. Solve ∣2x−5∣≤7∣2x−5∣≤7.
−1≤x≤6−1≤x≤6
The inequality ∣2x−5∣≤7∣2x−5∣≤7 is equivalent
to −7≤2x−5≤7−7≤2x−5≤7. Add 5 to all
parts: −2≤2x≤12−2≤2x≤12. Divide by
2: −1≤x≤6−1≤x≤6.
3. Solve the inequality x2−4x+3>0x2−4x+3>0.
x<1x<1 or x>3x>3

,Factorise: (x−1)(x−3)>0(x−1)(x−3)>0. The product
is positive when both factors are positive (x>3x>3)
or both are negative (x<1x<1). The critical points
are x=1x=1 and x=3x=3.
4. Solve 2x−2≥1x−22≥1.
2<x≤42<x≤4
Multiply
by (x−2)2>0(x−2)2>0: 2(x−2)≥(x−2)22(x−2)≥(x−2)
2. Expand: 2x−4≥x2−4x+42x−4≥x2−4x+4,
giving 0≥x2−6x+8=(x−2)(x−4)0≥x2−6x+8=(x−2)(x
−4). This holds for 2≤x≤42≤x≤4. Since x=2x=2 is
excluded (denominator zero), the solution
is 2<x≤42<x≤4.
5. Solve ∣3x+1∣>∣x−2∣∣3x+1∣>∣x−2∣.
x<−32x<−23 or x>14x>41
Square both sides: (3x+1)2>(x−2)2(3x+1)2>(x−2)2.
Expand: 9x2+6x+1>x2−4x+49x2+6x+1>x2−4x+4.
Simplify: 8x2+10x−3>08x2+10x−3>0.
Factorise: (2x+3)(4x−1)>0(2x+3)(4x−1)>0. Critical

, points are x=−32x=−23 and x=14x=41. The
quadratic is positive outside the roots.
6. Solve x2−1x+2≤0x+2x2−1≤0.
x≤−2x≤−2 or −1≤x≤1−1≤x≤1
Factorise
numerator: (x−1)(x+1)/(x+2)≤0(x−1)(x+1)/(x+2)≤0
. The critical points are x=−2,−1,1x=−2,−1,1. Use a
sign table: the expression is negative
for x<−2x<−2, positive for −2<x<−1−2<x<−1,
negative for −1<x<1−1<x<1, and positive
for x>1x>1. Include endpoints where the
expression is zero (x=±1x=±1). x=−2x=−2 is
excluded due to denominator zero, but the
inequality is ≤0≤0, so the solution
includes x<−2x<−2 (not x=−2x=−2)
and −1≤x≤1−1≤x≤1. However, checking the sign
table shows the expression is negative
for x<−2x<−2, so the solution
is x<−2x<−2 or −1≤x≤1−1≤x≤1. The answer given
in the prompt includes x≤−2x≤−2, which is
incorrect because at x=−2x=−2 the expression is