Edexcel C1 — Question 4 6 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeSolve equations with surds
DifficultyModerate -0.8 Part (a) is routine index law application with a mixed number. Part (b) requires rationalizing the denominator but follows a standard algebraic procedure with minimal steps. Both are straightforward C1-level exercises requiring recall and basic manipulation rather than problem-solving.
Spec1.02a Indices: laws of indices for rational exponents1.02b Surds: manipulation and rationalising denominators
4. (a) Evaluate \(\left( 5 \frac { 4 } { 9 } \right) ^ { - \frac { 1 } { 2 } }\).
(b) Find the value of \(x\) such that $$\frac { 1 + x } { x } = \sqrt { 3 } ,$$ giving your answer in the form \(a + b \sqrt { 3 }\) where \(a\) and \(b\) are rational.
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Notes
(a) \(\left(\frac{49}{9}\right)^{-\frac{1}{2}} = \sqrt{\frac{9}{49}} = \frac{3}{7}\)M1 A1
(b) \(1 + x = \sqrt{3}\, x\) → \(1 = x(\sqrt{3}-1)\)M1
\(x = \frac{1}{\sqrt{3}-1}\)A1
\(x = \frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{3}+1}{3-1} = \frac{1}{2} + \frac{1}{2}\sqrt{3}\)M1 A1 (6) 
## Question 4:

| Answer/Working | Marks | Notes |
|---|---|---|
| **(a)** $\left(\frac{49}{9}\right)^{-\frac{1}{2}} = \sqrt{\frac{9}{49}} = \frac{3}{7}$ | M1 A1 | |
| **(b)** $1 + x = \sqrt{3}\, x$ → $1 = x(\sqrt{3}-1)$ | M1 | |
| $x = \frac{1}{\sqrt{3}-1}$ | A1 | |
| $x = \frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{3}+1}{3-1} = \frac{1}{2} + \frac{1}{2}\sqrt{3}$ | M1 A1 | **(6)** |

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4. (a) Evaluate $\left( 5 \frac { 4 } { 9 } \right) ^ { - \frac { 1 } { 2 } }$.\\
(b) Find the value of $x$ such that

$$\frac { 1 + x } { x } = \sqrt { 3 } ,$$

giving your answer in the form $a + b \sqrt { 3 }$ where $a$ and $b$ are rational.\\

\hfill \mbox{\textit{Edexcel C1  Q4 [6]}}
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