Find the range of values of $x$ for which the binomial expansion below is valid.
$$\frac{3x-1}{(1-x)(2-3x)}$$
Initially I read this and thought, "ok so what can't $x$ be". You can't square root a negative and you can't divide by zero.
So $x$ can't be equal to $1$ because it makes the denominator $0$.
The same would happen for when $x=\frac{2}{3}$
These are the only values of $x$ that I can't get a valid output for therefore the range of values of $x$ is:
$$x\neq1$$
$$x\neq\frac{2}{3}$$
The answer in the Mark Scheme is:
$$-\frac{2}{3}<x<\frac{2}{3}$$
Huh?
$2$ is outside this range of values of $x$ and yet $f(2)$ gives a valid output.
What am I missing?
Best Answer
The binomial theorem for negative powers allows the expansion $$ (1-t)^{-1} = 1 + t + t^2 + t^3 + \cdots\tag1 $$ provided $|t|<1$. For your problem the first step is to rewrite your expression in a form that looks like the LHS of (1). To do this, write $$ {3x-1\over (1-x)(2-3x)} = \frac{-2}{1-x} + \frac3{2-3x}\tag2 $$ (you can use a partial fractions argument to see this). The first term of the RHS of (2) is now in the proper form; it can be expanded like (1) provided $|x|<1$. Now re-express the second term: $$\frac3{2-3x}=\frac32\frac1{(1-\frac32x)}\tag3$$ Therefore the second term can be expressed in a binomial expansion provided $|\frac32x|<1$, which is equivalent to $|x|<\frac23$. The overall expansion is therefore valid where both terms on the RHS of (2) can be validly expanded, which is $|x|<\frac23$.