OCR C3 2010 June — Question 7 10 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2010
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea between curve and line
DifficultyStandard +0.8 This is a multi-step integration problem requiring: (1) differentiation using chain rule to find the tangent equation, (2) finding where the tangent meets the x-axis, (3) calculating the area under the curve via integration with substitution, (4) calculating the triangular area, and (5) subtracting to find the shaded region. The integration of (3x-1)^4 requires substitution and careful handling of limits. While the individual techniques are standard C3 content, the multi-stage nature, need for exact answers, and geometric reasoning make this moderately challenging—above average but not exceptionally difficult.
Spec1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals

\includegraphics{figure_7} The diagram shows the curve with equation \(y = (3x - 1)^4\). The point P on the curve has coordinates \((1, 16)\) and the tangent to the curve at P meets the \(x\)-axis at the point Q. The shaded region is bounded by PQ, the \(x\)-axis and that part of the curve for which \(\frac{1}{3} \leqslant x \leqslant 1\). Find the exact area of this shaded region. [10]

AnswerMarks Guidance
Differentiate to obtain \(k_1(3x - 1)^3\)M1 any constant \(k_1\)
Obtain correct \(\left12(3x - 1)^3 \right \)
Substitute 1 to obtain 96A1
Attempt to find x-coordinate of \(Q\)M1 using tangent with \(y = 0\) or using gradient
Obtain \(\frac{2}{3}\)A1 or exact equiv
Integrate to obtain \(k_2(3x - 1)^5\)M1 any constant \(k_2\)
Obtain correct \(\frac{1}{4}(3x - 1)^5\)A1 or (unsimplified) equiv
Use limits \(\frac{1}{3}\) and 1 to obtain \(\frac{32}{15}\)A1
Attempt to find shaded area by correct processM1 integral – triangle or equiv
Obtain \(\left(\frac{32}{15} - \frac{1}{2} \times \frac{2}{3} \times 16\right)\) and hence \(\frac{4}{5}\)A1 or equiv
Differentiate to obtain $k_1(3x - 1)^3$ | M1 | any constant $k_1$

Obtain correct $\left| 12(3x - 1)^3 \right|$ | A1 | or (unsimplified) equiv

Substitute 1 to obtain 96 | A1 | —

Attempt to find x-coordinate of $Q$ | M1 | using tangent with $y = 0$ or using gradient

Obtain $\frac{2}{3}$ | A1 | or exact equiv

Integrate to obtain $k_2(3x - 1)^5$ | M1 | any constant $k_2$

Obtain correct $\frac{1}{4}(3x - 1)^5$ | A1 | or (unsimplified) equiv

Use limits $\frac{1}{3}$ and 1 to obtain $\frac{32}{15}$ | A1 | —

Attempt to find shaded area by correct process | M1 | integral – triangle or equiv

Obtain $\left(\frac{32}{15} - \frac{1}{2} \times \frac{2}{3} \times 16\right)$ and hence $\frac{4}{5}$ | A1 | or equiv

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\includegraphics{figure_7}

The diagram shows the curve with equation $y = (3x - 1)^4$. The point P on the curve has coordinates $(1, 16)$ and the tangent to the curve at P meets the $x$-axis at the point Q. The shaded region is bounded by PQ, the $x$-axis and that part of the curve for which $\frac{1}{3} \leqslant x \leqslant 1$. Find the exact area of this shaded region. [10]

\hfill \mbox{\textit{OCR C3 2010 Q7 [10]}}