| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Non-zero terms only |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard C4 binomial expansion and integration techniques. Part (i) requires routine manipulation to get the form (1+...)^n and applying the binomial theorem, part (ii) is term-by-term integration of a polynomial, and part (iii) is direct substitution into a given integral formula. All steps are procedural with no novel insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{\sqrt{4-x^2}} = 4^{-\frac{1}{2}}\left(1-\frac{1}{4}x^2\right)^{-\frac{1}{2}}\) | M1 | |
| \(= \frac{1}{2}\left[1+\left(-\frac{1}{2}\right)\left(-\frac{1}{4}x^2\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{1}{4}x^2\right)^2+\ldots\right]\) | M1 | Binomial coefficients correct |
| A1 | Complete correct expression inside bracket | |
| \(= \frac{1}{2}+\frac{1}{16}x^2+\frac{3}{256}x^4+\ldots\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^1 \frac{1}{\sqrt{4-x^2}}\,dx \approx \int_0^1\left(\frac{1}{2}+\frac{1}{16}x^2+\frac{3}{256}x^4\right)dx\) | M1ft | |
| \(= \left[\frac{1}{2}x+\frac{1}{48}x^3+\frac{3}{1280}x^5\right]_0^1\) | ||
| \(= \frac{1}{2}+\frac{1}{48}+\frac{3}{1280} = 0.5232\) (to 4 s.f.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^1 \frac{1}{\sqrt{4-x^2}}\,dx = \left[\arcsin\frac{x}{2}\right]_0^1 = \frac{\pi}{6} = 0.5236\) | B1 | |
| [7] |
## Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{4-x^2}} = 4^{-\frac{1}{2}}\left(1-\frac{1}{4}x^2\right)^{-\frac{1}{2}}$ | M1 | |
| $= \frac{1}{2}\left[1+\left(-\frac{1}{2}\right)\left(-\frac{1}{4}x^2\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{1}{4}x^2\right)^2+\ldots\right]$ | M1 | Binomial coefficients correct |
| | A1 | Complete correct expression inside bracket |
| $= \frac{1}{2}+\frac{1}{16}x^2+\frac{3}{256}x^4+\ldots$ | A1 | cao |
## Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^1 \frac{1}{\sqrt{4-x^2}}\,dx \approx \int_0^1\left(\frac{1}{2}+\frac{1}{16}x^2+\frac{3}{256}x^4\right)dx$ | M1ft | |
| $= \left[\frac{1}{2}x+\frac{1}{48}x^3+\frac{3}{1280}x^5\right]_0^1$ | | |
| $= \frac{1}{2}+\frac{1}{48}+\frac{3}{1280} = 0.5232$ (to 4 s.f.) | A1 | |
## Question 7(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^1 \frac{1}{\sqrt{4-x^2}}\,dx = \left[\arcsin\frac{x}{2}\right]_0^1 = \frac{\pi}{6} = 0.5236$ | B1 | |
| **[7]** | | |7 (i) Find the first three non-zero terms of the binomial expansion of $\frac { 1 } { \sqrt { 4 - x ^ { 2 } } }$ for $| x | < 2$. [4]\\
(ii) Use this result to find an approximation for $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x$, rounding your answer to\\
4 significant figures.\\
(iii) Given that $\int \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x = \arcsin \left( \frac { 1 } { 2 } x \right) + c$, evaluate $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x$, rounding your answer to 4 significant figures.
\hfill \mbox{\textit{OCR MEI C4 Q7 [7]}}