Edexcel C12 2017 October — Question 9 10 marks

Exam BoardEdexcel
ModuleC12 (Core Mathematics 1 & 2)
Year2017
SessionOctober
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeDeduce related integral from numerical approximation
DifficultyModerate -0.8 This is a straightforward multi-part question testing basic skills: sketching an exponential graph (routine), applying the trapezium rule with given values (mechanical calculation), and using integral properties (linearity and substitution) to deduce related values. All parts are standard textbook exercises requiring no problem-solving insight, making it easier than average for A-level.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.09f Trapezium rule: numerical integration
  1. (a) Given that \(a\) is a constant, \(a > 1\), sketch the graph of
$$y = a ^ { x } , \quad x \in \mathbb { R }$$ On your diagram show the coordinates of the point where the graph crosses the \(y\)-axis.
(2) The table below shows corresponding values of \(x\) and \(y\) for \(y = 2 ^ { x }\)
\(x\)- 4- 2024
\(y\)0.06250.251416
(b) Use the trapezium rule, with all of the values of \(y\) from the table, to find an approximate value, to 2 decimal places, for $$\int _ { - 4 } ^ { 4 } 2 ^ { x } \mathrm {~d} x$$ (c) Use the answer to part (b) to find an approximate value for
  1. \(\int _ { - 4 } ^ { 4 } 2 ^ { x + 2 } \mathrm {~d} x\)
  2. \(\int _ { - 4 } ^ { 4 } \left( 3 + 2 ^ { x } \right) \mathrm { d } x\)
    \includegraphics[max width=\textwidth, alt={}, center]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-23_86_47_2617_1886}
Question 9:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct shape or \(y\)-intercept at 1B1 Curve must lie only in quadrants 1 and 2, positive and increasing gradient left to right, gradient approximately 0 at left end; condone curve appearing straight on rhs
Fully correct shape and interceptB1 Gradient must appear ~0 at lh end and reach level more than halfway below intercept at \((0,1)\); allow \(x=0, y=1\) in text
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
State \(h=2\), or use of \(\frac{1}{2}\times2\)B1 Strip width of 2
\(\{0.0625+16+2(0.25+1+4)\}\)M1, A1 M1: correct outer bracket structure with first \(y\) value + last \(y\) value + 2×(sum of remaining \(y\) values), no additional values; A1: correct bracket \(\{......\}\)
\(\frac{1}{2}\times2\times\{26.5625\} = \text{awrt } 26.56\)A1cao Exact answer \(= \frac{425}{16}\); accept separate trapezia: B1 for \(h=1\), M1 for \(\frac{1}{2}h(a+b)\) used 3 or 4 times, A1 if all correct
Part (c)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(4\times(b) = \text{awrt } 106\)M1, A1ft M1: attempt at \(4\times(b)\); also allow repeating trapezium rule with each value \(\times4\); A1ft: awrt 106 or ft on \(4\times(b)\); exact answer \(= \frac{425}{4}\)
Part (c)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(24+(b) = \text{awrt } 50.6\)M1, A1ft M1: attempt at \(24+(b)\) or \([3x]_{-4}^{4}+(b)\); also allow repeating trapezium rule with each value \(+3\); A1ft: awrt 50.6 or ft on \(24+(b)\); exact answer \(= \frac{809}{16}\) 
## Question 9:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape or $y$-intercept at 1 | B1 | Curve must lie only in quadrants 1 and 2, positive and increasing gradient left to right, gradient approximately 0 at left end; condone curve appearing straight on rhs |
| Fully correct shape and intercept | B1 | Gradient must appear ~0 at lh end and reach level more than halfway below intercept at $(0,1)$; allow $x=0, y=1$ in text |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| State $h=2$, or use of $\frac{1}{2}\times2$ | B1 | Strip width of 2 |
| $\{0.0625+16+2(0.25+1+4)\}$ | M1, A1 | M1: correct outer bracket structure with first $y$ value + last $y$ value + 2×(sum of remaining $y$ values), no additional values; A1: correct bracket $\{......\}$ |
| $\frac{1}{2}\times2\times\{26.5625\} = \text{awrt } 26.56$ | A1cao | Exact answer $= \frac{425}{16}$; accept separate trapezia: B1 for $h=1$, M1 for $\frac{1}{2}h(a+b)$ used 3 or 4 times, A1 if all correct |

### Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\times(b) = \text{awrt } 106$ | M1, A1ft | M1: attempt at $4\times(b)$; also allow repeating trapezium rule with each value $\times4$; A1ft: awrt 106 or ft on $4\times(b)$; exact answer $= \frac{425}{4}$ |

### Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $24+(b) = \text{awrt } 50.6$ | M1, A1ft | M1: attempt at $24+(b)$ or $[3x]_{-4}^{4}+(b)$; also allow repeating trapezium rule with each value $+3$; A1ft: awrt 50.6 or ft on $24+(b)$; exact answer $= \frac{809}{16}$ |
\begin{enumerate}
  \item (a) Given that $a$ is a constant, $a > 1$, sketch the graph of
\end{enumerate}

$$y = a ^ { x } , \quad x \in \mathbb { R }$$

On your diagram show the coordinates of the point where the graph crosses the $y$-axis.\\
(2)

The table below shows corresponding values of $x$ and $y$ for $y = 2 ^ { x }$

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 4 & - 2 & 0 & 2 & 4 \\
\hline
$y$ & 0.0625 & 0.25 & 1 & 4 & 16 \\
\hline
\end{tabular}
\end{center}

(b) Use the trapezium rule, with all of the values of $y$ from the table, to find an approximate value, to 2 decimal places, for

$$\int _ { - 4 } ^ { 4 } 2 ^ { x } \mathrm {~d} x$$

(c) Use the answer to part (b) to find an approximate value for\\
(i) $\int _ { - 4 } ^ { 4 } 2 ^ { x + 2 } \mathrm {~d} x$\\
(ii) $\int _ { - 4 } ^ { 4 } \left( 3 + 2 ^ { x } \right) \mathrm { d } x$

\begin{center}

\end{center}

\includegraphics[max width=\textwidth, alt={}, center]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-23_86_47_2617_1886}\\

\hfill \mbox{\textit{Edexcel C12 2017 Q9 [10]}}
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