Challenging +1.2 This is a standard Further Maths question on using rectangles to bound integrals. Students must recognize that the sum of rectangle areas (upper/lower bounds) relates to the integral of 1/√(x²+2x), then integrate using a standard substitution or recognition of d/dx[ln(x+√(x²+2x))]. While it requires multiple steps and careful algebraic manipulation, the technique is well-practiced in Further Maths syllabi and follows a predictable pattern. The integration itself is moderately challenging but recognizable to prepared students.
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\includegraphics[max width=\textwidth, alt={}, center]{323ac7a5-4690-441d-87fc-325a393098fa-10_585_1349_258_358}
The diagram shows the curve \(\mathrm { y } = \frac { 1 } { \sqrt { \mathrm { x } ^ { 2 } + 2 \mathrm { x } } }\) for \(x > 0\), together with a set of \(( n - 1 )\) rectangles of unit
width.
By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { \sqrt { r ^ { 2 } + 2 r } } < \ln \left( n + 1 + \sqrt { n ^ { 2 } + 2 n } \right) + \frac { 1 } { 3 } \sqrt { 3 } - \ln ( 2 + \sqrt { 3 } )$$
Forms sum of areas of rectangles. At least three terms, including first and last for A1. SC: Areas of these three rectangles written down but sum not clearly formed scores B1
## Question 6:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Sum}=\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{15}}+\cdots+\frac{1}{\sqrt{n^2+2n}}=\sum_{r=2}^{n}\frac{1}{\sqrt{r^2+2r}}$ | **M1 A1** | Forms sum of areas of rectangles. At least three terms, including first and last for A1. SC: Areas of these three rectangles written down but sum not clearly formed scores B1 |
| $<\int_1^n\frac{1}{\sqrt{x^2+2x}}\,dx$ | **M1** | Compares with integral |
| $x^2+2x=(x+1)^2-1$ | **B1** | Completes the square |
| $\int_1^n\frac{1}{\sqrt{(x+1)^2-1}}\,dx=\left[\cosh^{-1}(x+1)\right]_1^n$ | **M1 A1** | Uses formula. Correct limits for A1 |
| $\left[\ln\!\left(x+1+\sqrt{x^2+2x}\right)\right]_1^n$ | **M1 A1** | Uses logarithmic form of $\cosh^{-1}$ |
| $\ln(n+1+\sqrt{n^2+2n})-\ln(2+\sqrt{3})$ | **M1** | Inserts limits |
| $\sum_{r=1}^{n}\frac{1}{\sqrt{r^2+2r}}<\ln(n+1+\sqrt{n^2+2n})-\ln(2+\sqrt{3})+\frac{1}{3}\sqrt{3}$ | **A1** | Adds term from $r=1$. AG |
| | **10** | |
6\\
\includegraphics[max width=\textwidth, alt={}, center]{323ac7a5-4690-441d-87fc-325a393098fa-10_585_1349_258_358}
The diagram shows the curve $\mathrm { y } = \frac { 1 } { \sqrt { \mathrm { x } ^ { 2 } + 2 \mathrm { x } } }$ for $x > 0$, together with a set of $( n - 1 )$ rectangles of unit\\
width.
By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { \sqrt { r ^ { 2 } + 2 r } } < \ln \left( n + 1 + \sqrt { n ^ { 2 } + 2 n } \right) + \frac { 1 } { 3 } \sqrt { 3 } - \ln ( 2 + \sqrt { 3 } )$$
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q6 [10]}}