OCR C1 2007 January — Question 4 5 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2007
SessionJanuary
Marks5
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Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeSolve power equations
DifficultyModerate -0.5 This is a standard substitution problem where letting y = x^(1/3) transforms it into the quadratic y² + 3y - 10 = 0, which factors easily to (y+5)(y-2) = 0. While it requires recognizing the substitution pattern and cubing the solutions, it's a routine C1 exercise with straightforward algebraic manipulation, making it slightly easier than average.
Spec1.02f Solve quadratic equations: including in a function of unknown
4 Solve the equation \(x ^ { \frac { 2 } { 3 } } + 3 x ^ { \frac { 1 } { 3 } } - 10 = 0\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Let \(y = x^{\frac{1}{3}}\)*M1 Attempt a substitution to obtain a quadratic or factorise with \(\sqrt[3]{x}\) in each bracket
\(y^2 + 3y - 10 = 0\)
\((y-2)(y+5) = 0\)DM1 Correct attempt to solve quadratic
\(y = 2,\ y = -5\)A1 Both values correct
\(x = 2^3,\ x = (-5)^3\)DM1 Attempt cube
\(x = 8,\ x = -125\)A1 ft [5] Both answers correctly followed through. SR B2 \(x=8\) from T&I 
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $y = x^{\frac{1}{3}}$ | *M1 | Attempt a substitution to obtain a quadratic or factorise with $\sqrt[3]{x}$ in each bracket |
| $y^2 + 3y - 10 = 0$ | | |
| $(y-2)(y+5) = 0$ | DM1 | Correct attempt to solve quadratic |
| $y = 2,\ y = -5$ | A1 | Both values correct |
| $x = 2^3,\ x = (-5)^3$ | DM1 | Attempt cube |
| $x = 8,\ x = -125$ | A1 ft [5] | Both answers correctly followed through. **SR** B2 $x=8$ from T&I |

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4 Solve the equation $x ^ { \frac { 2 } { 3 } } + 3 x ^ { \frac { 1 } { 3 } } - 10 = 0$.

\hfill \mbox{\textit{OCR C1 2007 Q4 [5]}}
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