| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Direct single expansion substitution |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion with fractional powers. Part (i) is simple algebraic verification, part (ii) requires routine application of (1+x)^n formula with n=-1/2, part (iii) is direct substitution, and part (iv) is basic percentage error calculation. All steps are standard C4 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks |
|---|---|
| (i) \(f(\frac{1}{10}) = \frac{3}{\sqrt{1-\frac{1}{10}}} = \frac{3}{\sqrt{\frac{9}{10}}} = \frac{3}{(\frac{3}{\sqrt{10}})} = \sqrt{10}\) | M1 A1 |
| (ii) \(3(1-x)^{-\frac{1}{2}} = 3[1 + (-\frac{1}{2})(-x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(-x)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3×2}(-x)^3 + ...]\) | M1 |
| \(= 3 + \frac{3}{2}x + \frac{9}{8}x^2 + \frac{15}{16}x^3 + ...\) | A3 |
| (iii) \(\sqrt{10} = f(\frac{1}{10}) = 3 + \frac{3}{20} + \frac{9}{800} + \frac{15}{16000} = 3.1621875 (8sf)\) | B1 |
| (iv) \(\frac{\sqrt{10} - 3.1621875}{\sqrt{10}} × 100\% = 0.003\% (1sf)\) | M1 A1 |
**(i)** $f(\frac{1}{10}) = \frac{3}{\sqrt{1-\frac{1}{10}}} = \frac{3}{\sqrt{\frac{9}{10}}} = \frac{3}{(\frac{3}{\sqrt{10}})} = \sqrt{10}$ | M1 A1
**(ii)** $3(1-x)^{-\frac{1}{2}} = 3[1 + (-\frac{1}{2})(-x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(-x)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3×2}(-x)^3 + ...]$ | M1
$= 3 + \frac{3}{2}x + \frac{9}{8}x^2 + \frac{15}{16}x^3 + ...$ | A3
**(iii)** $\sqrt{10} = f(\frac{1}{10}) = 3 + \frac{3}{20} + \frac{9}{800} + \frac{15}{16000} = 3.1621875 (8sf)$ | B1
**(iv)** $\frac{\sqrt{10} - 3.1621875}{\sqrt{10}} × 100\% = 0.003\% (1sf)$ | M1 A16.
$$f ( x ) = \frac { 3 } { \sqrt { 1 - x } } , | x | < 1$$
(i) Show that $\mathrm { f } \left( \frac { 1 } { 10 } \right) = \sqrt { 10 }$.\\
(ii) Expand $\mathrm { f } ( x )$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, simplifying each coefficient.\\
(iii) Use your expansion to find an approximate value for $\sqrt { 10 }$, giving your answer to 8 significant figures.\\
(iv) Find, to 1 significant figure, the percentage error in your answer to part (c).\\
\hfill \mbox{\textit{OCR C4 Q6 [9]}}