CAIE P1 2018 November — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks3
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Mark schemeDownload PDF ↗
TopicSolving quadratics and applications
TypeQuadratic in x^(1/2) - substitution u = √x
DifficultyModerate -0.5 This is a straightforward substitution question where students let u = x^(1/2) to obtain a quadratic in u, then solve and back-substitute. It requires only standard technique with no conceptual challenges, making it slightly easier than average, though the fractional power and need for two valid solutions prevents it from being trivial.
Spec1.02f Solve quadratic equations: including in a function of unknown
1 Showing all necessary working, solve the equation \(4 x - 11 x ^ { \frac { 1 } { 2 } } + 6 = 0\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\((4x^{1/2}-3)(x^{1/2}-2)\) oe soi; Alt: \(4x+6=11\sqrt{x} \Rightarrow 16x^2-73x+36\)M1 Attempt solution for \(x^{1/2}\) or sub \(u=x^{1/2}\)
\(x^{1/2}=3/4\) or \(2\); \((16x-9)(x-4)\)A1 Reasonable solutions for \(x^{1/2}\) implies M1 (\(x=2, 3/4\), M1A0)
\(x=9/16\) oe or \(4\)A1 Little or no working shown scores SCB3, spotting one solution, B0
Total: 3 
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(4x^{1/2}-3)(x^{1/2}-2)$ oe soi; Alt: $4x+6=11\sqrt{x} \Rightarrow 16x^2-73x+36$ | M1 | Attempt solution for $x^{1/2}$ or sub $u=x^{1/2}$ |
| $x^{1/2}=3/4$ or $2$; $(16x-9)(x-4)$ | A1 | Reasonable solutions for $x^{1/2}$ implies M1 ($x=2, 3/4$, M1A0) |
| $x=9/16$ oe or $4$ | A1 | Little or no working shown scores SCB3, spotting one solution, B0 |
| **Total: 3** | | |

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1 Showing all necessary working, solve the equation $4 x - 11 x ^ { \frac { 1 } { 2 } } + 6 = 0$.\\

\hfill \mbox{\textit{CAIE P1 2018 Q1 [3]}}
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