9709 P31 - Jun 2015 - Q3
2023
Show that, for small values of \(x^2\),
\((1 - 2x^2)^{-2} - (1 + 6x^2)^{\frac{2}{3}} \approx kx^4\),
where the value of the constant \(k\) is to be determined.
Solution
To solve this, we need to expand both expressions for small values of \(x^2\) and find the coefficient of \(x^4\).
First, consider \((1 - 2x^2)^{-2}\). Using the binomial expansion for \((1 - u)^{-n}\), we have:
\((1 - 2x^2)^{-2} = 1 + 4x^2 + 12x^4 + \cdots\)
Next, consider \((1 + 6x^2)^{\frac{2}{3}}\). Using the binomial expansion for \((1 + u)^{n}\), we have:
\((1 + 6x^2)^{\frac{2}{3}} = 1 + 4x^2 - 4x^4 + \cdots\)
Subtracting these expansions:
\((1 - 2x^2)^{-2} - (1 + 6x^2)^{\frac{2}{3}} = (1 + 4x^2 + 12x^4) - (1 + 4x^2 - 4x^4)\)
\(= 12x^4 + 4x^4 = 16x^4\)
Thus, \(k = 16\).
