Section A: Pure Mathematics 1 (a) Show that $\displaystyle \frac{1}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{1}{2x^{2}}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right).$ (b) Hence use appropriate expansions from the List of Formulae (MF26) to find the first two non-zero terms in the series expansion of $\displaystyle \frac{x^{2}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}},\; x\ne 0$ in ascending powers of $x$. (c) State the set of values of $x$ for which the series expansion is valid. (d) It is given that the two terms found in part (b) are equal to the first two terms in the series expansion of $\cos\left(a x^{b}\right)$. Find the possible value(s) of the constants $a$ and $b$.

1. Multiply the numerator and denominator by the conjugate of the denominator to rationalise it.

$\frac{1}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \times \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$

2. Simplify the denominator using the difference of two squares identity to arrive at the required result.

$= \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{(1+x^{2}) - (1-x^{2})}$
$= \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{2x^{2}}$
$= \frac{1}{2x^{2}}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right) \quad \text{(shown)}$

1. Use the result from part (a) to simplify the given expression.

$\frac{x^{2}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} = x^2 \left[ \frac{1}{2x^2}\left(\sqrt{1+x^{2}}+\sqrt{1-x^{2}}\right) \right]$
$= \frac{1}{2} \left( (1+x^2)^{\frac{1}{2}} + (1-x^2)^{\frac{1}{2}} \right)$

2. Expand $(1+x^2)^{\frac{1}{2}}$ and $(1-x^2)^{\frac{1}{2}}$ using the standard binomial series expansion up to the $x^4$ term.

Using $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$
$(1+x^2)^{\frac{1}{2}} = 1 + \frac{1}{2}(x^2) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x^2)^2 + \dots = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$
$(1-x^2)^{\frac{1}{2}} = 1 + \frac{1}{2}(-x^2) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(-x^2)^2 + \dots = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$

3. Substitute the expansions back into the simplified expression and collect terms to find the first two non-zero terms.

$\frac{1}{2} \left( (1+x^2)^{\frac{1}{2}} + (1-x^2)^{\frac{1}{2}} \right)$
$= \frac{1}{2} \left[ \left( 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 \right) + \left( 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 \right) \right]$
$= \frac{1}{2} \left( 2 - \frac{1}{4}x^4 \right)$
$= 1 - \frac{1}{8}x^4$

1. State the validity condition for the binomial expansions and combine it with the given domain restriction.

The series expansions used in part (b) are for $(1+x^2)^{\frac{1}{2}}$ and $(1-x^2)^{\frac{1}{2}}$.
By the standard binomial expansion, these are valid when $|x^2| < 1$ and $|-x^2| < 1$.
$x^2 < 1 \implies -1 < x < 1$
Given the restriction $x \ne 0$ for the original expression in part (b), we must exclude $0$.
Therefore, the set of values is $-1 < x < 1, \; x \ne 0$.

1. Determine the result from part (b) (as it is required to solve this part).

From part (a), the expression simplifies to $\frac{1}{2}\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)$.
Using the standard binomial expansion:
$(1+x^2)^{1/2} = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$
$(1-x^2)^{1/2} = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 + \dots$
Adding them and multiplying by $\frac{1}{2}$, the first two non-zero terms from part (b) are:
$\frac{1}{2}\left(2 - \frac{1}{4}x^4\right) = 1 - \frac{1}{8}x^4$

2. Expand $\cos(ax^b)$ using the standard series.

Using the standard series expansion $\cos u \approx 1 - \frac{u^2}{2!}$, substitute $u = ax^b$:
$\cos(ax^b) \approx 1 - \frac{(ax^b)^2}{2} = 1 - \frac{a^2 x^{2b}}{2}$

3. Equate the expansions to find the possible values of $a$ and $b$.

Equating the first two terms of both expansions:
$1 - \frac{a^2 x^{2b}}{2} \equiv 1 - \frac{1}{8}x^4$

Comparing the powers of $x$:
$2b = 4 \implies b = 2$

Comparing the coefficients of the second term:
$-\frac{a^2}{2} = -\frac{1}{8}$
$a^2 = \frac{1}{4} \implies a = \pm \frac{1}{2}$