CAIE P1 2021 March — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2021
SessionMarch
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSolving quadratics and applications
TypeSubstitution to solve disguised quadratic
DifficultyStandard +0.3 This is a standard substitution question where students let u = (2x-3)² to obtain a quadratic in u, then solve and back-substitute. While it requires recognizing the substitution pattern and handling the reciprocal term, it's a routine technique taught explicitly in P1 with straightforward algebraic manipulation. Slightly easier than average due to its mechanical nature.
Spec1.02c Simultaneous equations: two variables by elimination and substitution
2 By using a suitable substitution, solve the equation $$( 2 x - 3 ) ^ { 2 } - \frac { 4 } { ( 2 x - 3 ) ^ { 2 } } - 3 = 0$$
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(u = 2x - 3\) leading to \(u^4 - 3u^2 - 4\ [= 0]\)M1 Or \(u = (2x-3)^2\) leading to \(u^2 - 3u - 4\ [=0]\)
\((u^2 - 4)(u^2 + 1)\ [= 0]\)M1 Or \((u-4)(u+1)\ [=0]\)
\(2x - 3 = [\pm]\, 2\)A1
\(x = \dfrac{1}{2},\ \dfrac{5}{2}\) onlyA1
Total: 4 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = 2x - 3$ leading to $u^4 - 3u^2 - 4\ [= 0]$ | M1 | Or $u = (2x-3)^2$ leading to $u^2 - 3u - 4\ [=0]$ |
| $(u^2 - 4)(u^2 + 1)\ [= 0]$ | M1 | Or $(u-4)(u+1)\ [=0]$ |
| $2x - 3 = [\pm]\, 2$ | A1 | |
| $x = \dfrac{1}{2},\ \dfrac{5}{2}$ **only** | A1 | |
| | **Total: 4** | |
2 By using a suitable substitution, solve the equation

$$( 2 x - 3 ) ^ { 2 } - \frac { 4 } { ( 2 x - 3 ) ^ { 2 } } - 3 = 0$$

\hfill \mbox{\textit{CAIE P1 2021 Q2 [4]}}
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Q1 5 Q2 4 Q3 4 Q4 5 Q5 6 Q6 7 Q7 8 Q8 7 Q9 9 Q10 8 Q11 12