OCR PURE — Question 8 9 marks

Exam BoardOCR
ModulePURE
Marks9
PaperDownload PDF ↗
TopicAreas by integration
TypeArea involving fractional powers
DifficultyStandard +0.3 This is a straightforward integration problem requiring students to set up and solve ∫(2x^(1/3) - 7x^(-1/3))dx = 45 between limits 8 and a. The integration is routine (power rule with fractional indices), and solving the resulting equation for 'a' involves basic algebra. While it requires careful manipulation of surds and fractional powers, it's a standard textbook exercise with no novel problem-solving required, making it slightly easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals
8 In this question you must show detailed reasoning. The diagram shows part of the graph of \(y = 2 x ^ { \frac { 1 } { 3 } } - \frac { 7 } { x ^ { \frac { 1 } { 3 } } }\). The shaded region is enclosed by the curve, the \(x\)-axis and the lines \(x = 8\) and \(x = a\), where \(a > 8\). \includegraphics[max width=\textwidth, alt={}, center]{efde7b10-b4f3-469f-ba91-b765a16ea835-5_577_1164_477_438} Given that the area of the shaded region is 45 square units, find the value of \(a\).
Question 8:
\(\int_8^a 2x^{\frac{1}{3}} - 7x^{-\frac{1}{3}}dx = 45\)
AnswerMarks Guidance
\(\left[\frac{2x^{\frac{4}{3}}}{\left(\frac{4}{3}\right)} - \frac{7x^{\frac{2}{3}}}{\left(\frac{2}{3}\right)}\right]_8^a (= 45)\)M1\* AO3.1a
A1AO1.1 A1 – 1 term correct (accept unsimplified)
A1AO1.1 A1 – both correct (accept unsimplified)
\(\frac{3}{2}a^{\frac{4}{3}} - \frac{21}{2}a^{\frac{2}{3}} - \left(\frac{3}{2}(8)^{\frac{4}{3}} - \frac{21}{2}(8)^{\frac{2}{3}}\right)(= 45)\)Dep\*M1 AO1.1
\(\frac{3}{2}a^{\frac{4}{3}} - \frac{21}{2}a^{\frac{2}{3}} - (24 - 42)(= 45)\)A1 AO1.1
M1AO1.1 Equate integrated expression to 45 – dependent on both previous M marks
\(a^{\frac{4}{3}} - 7a^{\frac{2}{3}} - 18 = 0\)
AnswerMarks Guidance
\(\left(a^{\frac{2}{3}} - 9\right)\left(a^{\frac{2}{3}} + 2\right) = 0\)M1 AO3.1a
\(a^{\frac{2}{3}} = 9\) (and \(a^{\frac{2}{3}} = -2\))A1 AO1.1
\(a = 27\) onlyA1 AO2.2a 
## Question 8:

$\int_8^a 2x^{\frac{1}{3}} - 7x^{-\frac{1}{3}}dx = 45$

$\left[\frac{2x^{\frac{4}{3}}}{\left(\frac{4}{3}\right)} - \frac{7x^{\frac{2}{3}}}{\left(\frac{2}{3}\right)}\right]_8^a (= 45)$ | **M1\*** | AO3.1a | M1 – attempt integration (increase in power by 1 for at least 1 term)

| **A1** | AO1.1 | A1 – 1 term correct (accept unsimplified)

| **A1** | AO1.1 | A1 – both correct (accept unsimplified)

$\frac{3}{2}a^{\frac{4}{3}} - \frac{21}{2}a^{\frac{2}{3}} - \left(\frac{3}{2}(8)^{\frac{4}{3}} - \frac{21}{2}(8)^{\frac{2}{3}}\right)(= 45)$ | **Dep\*M1** | AO1.1 | $F(a) - F(8)$

$\frac{3}{2}a^{\frac{4}{3}} - \frac{21}{2}a^{\frac{2}{3}} - (24 - 42)(= 45)$ | **A1** | AO1.1 | oe

| **M1** | AO1.1 | Equate integrated expression to 45 – dependent on both previous M marks

$a^{\frac{4}{3}} - 7a^{\frac{2}{3}} - 18 = 0$

$\left(a^{\frac{2}{3}} - 9\right)\left(a^{\frac{2}{3}} + 2\right) = 0$ | **M1** | AO3.1a | Attempt to solve quadratic in $a^{\frac{2}{3}}$ | SC if M0 for fourth M mark then award

$a^{\frac{2}{3}} = 9$ (and $a^{\frac{2}{3}} = -2$) | **A1** | AO1.1 | B1 for $a^{\frac{2}{3}} = 9$

$a = 27$ only | **A1** | AO2.2a | B1 $a = 27$ only | If $a = 27$ with no working then 0/9

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8 In this question you must show detailed reasoning.
The diagram shows part of the graph of $y = 2 x ^ { \frac { 1 } { 3 } } - \frac { 7 } { x ^ { \frac { 1 } { 3 } } }$. The shaded region is enclosed by the curve, the $x$-axis and the lines $x = 8$ and $x = a$, where $a > 8$.\\
\includegraphics[max width=\textwidth, alt={}, center]{efde7b10-b4f3-469f-ba91-b765a16ea835-5_577_1164_477_438}

Given that the area of the shaded region is 45 square units, find the value of $a$.

\hfill \mbox{\textit{OCR PURE  Q8 [9]}}
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