| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule with stated number of strips |
| Difficulty | Moderate -0.3 This is a comprehensive multi-part question covering standard C2 techniques (integration of powers, tangent equations, trapezium rule) with straightforward applications. While it has many parts (7 marks total), each individual step is routine: converting to index form, basic integration, finding a tangent line, and applying the trapezium rule formula. The transformation in part (c) is simple horizontal translation. Slightly easier than average due to the scaffolded nature and lack of problem-solving insight required. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a)(i) | \(\sqrt{x} = x^{\frac{1}{2}}\) | B1 |
| 4(a)(ii) | \(\int \sqrt{x}\,dx = \frac{x^{1.5}}{1.5} \{+c\}\) | M1, A1 ft |
| 4(a)(iii) | Area = \(\int_1^4 \sqrt{x}\,dx\) = \(\frac{4^{1.5}}{1.5} - \frac{1}{1.5}\) = \(\frac{14}{3}\) | B1, M1, A1 |
| 4(b)(i) | \(y = x^2 \Rightarrow \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\); When x= 4, y'(4) = 0.25; When x = 4, y = 2; Equation of tangent: \(y - 2 = \frac{1}{4}(x - 4)\) | M1, M1, B1, A1 |
| 4(b)(ii) | When x = 0, y = 1 B(0, 1); When y = 0, x = −4 A(−4, 0); Area = 0.5(1)(4) = 2 | M1, A1 ft, A1 ft |
| 4(c) | Translation; \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) | B1, B1 |
| 4(d) | h = 1; Integral = h/2 {......}; {....} = f(1) + 2[f(2) + f(3)] + f(4); {....} = 0 + 2[1 + \(\sqrt{2}\)] + \(\sqrt{3}\); Integral = \(\frac{1}{2}\{2(1+1.414...) +1.732\}\); Integral = 0.5× 6.560..., = 3.28 to 3sf | B1, M1, A1, A1 |
4(a)(i) | $\sqrt{x} = x^{\frac{1}{2}}$ | B1 | Accept p = 0.5 |
4(a)(ii) | $\int \sqrt{x}\,dx = \frac{x^{1.5}}{1.5} \{+c\}$ | M1, A1 ft | Index raised by 1; Correct ft on p. Condone missing '+c' |
4(a)(iii) | Area = $\int_1^4 \sqrt{x}\,dx$ = $\frac{4^{1.5}}{1.5} - \frac{1}{1.5}$ = $\frac{14}{3}$ | B1, M1, A1 | Limits 1 and 4 PI; F(4) − F(1); Accept 4.66 or better |
4(b)(i) | $y = x^2 \Rightarrow \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$; When x= 4, y'(4) = 0.25; When x = 4, y = 2; Equation of tangent: $y - 2 = \frac{1}{4}(x - 4)$ | M1, M1, B1, A1 | Index $(p-1)$ ft; Attempt to find y'(4); accept other forms |
4(b)(ii) | When x = 0, y = 1 B(0, 1); When y = 0, x = −4 A(−4, 0); Area = 0.5(1)(4) = 2 | M1, A1 ft, A1 ft | Subs x = 0 and then y = 0 into equation; of tangent. PI Correct ft $y_a$ and $x_a$ (may be awarded as part of area calculation); ft wrong sloping tangent and max of 1 further slip. Final answer must be +'ve |
4(c) | Translation; $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ | B1, B1 | 'Translation'/'translate(d)'; Accept equivalent in words provided linked to 'translation/move/shift' (Note: B0B1 is possible) |
4(d) | h = 1; Integral = h/2 {......}; {....} = f(1) + 2[f(2) + f(3)] + f(4); {....} = 0 + 2[1 + $\sqrt{2}$] + $\sqrt{3}$; Integral = $\frac{1}{2}\{2(1+1.414...) +1.732\}$; Integral = 0.5× 6.560..., = 3.28 to 3sf | B1, M1, A1, A1 | PI; OE summing of areas of the three traps; Condone 1 numerical slip; CAO Must be 3.28 |4 The diagram shows a curve $C$ with equation $y = \sqrt { x }$. The point $O$ is the origin $( 0,0 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{37627fc4-a90b-4f3b-9b10-0a9e200f8485-3_488_1136_1009_443}
The region bounded by the curve $C$, the $x$-axis and the vertical lines $x = 1$ and $x = 4$ is shown shaded in the diagram.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write $\sqrt { x }$ in the form $x ^ { p }$, where $p$ is a constant.
\item Find $\int \sqrt { x } \mathrm {~d} x$.
\item Hence find the area of the shaded region.
\end{enumerate}\item The point on $C$ for which $x = 4$ is $P$. The tangent to $C$ at the point $P$ intersects the $x$-axis and the $y$-axis at the points $A$ and $B$ respectively.
\begin{enumerate}[label=(\roman*)]
\item Find an equation for the tangent to the curve $C$ at the point $P$.
\item Find the area of the triangle $A O B$.
\end{enumerate}\item Describe the single geometrical transformation by which the curve with equation $y = \sqrt { x - 1 }$ can be obtained from the curve $C$.
\item Use the trapezium rule with four ordinates (three strips) to find an approximation for $\int _ { 1 } ^ { 4 } \sqrt { x - 1 } \mathrm {~d} x$, giving your answer to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q4 [19]}}