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Jaxxnuts · Jul 28, 2017

1. For every pair of numbers a&b, the function f(x) satisfies b^2 f(a)=a^2 f(b)
If f(2) is not equal to 0 find
((f(5)-f(1))/2
2. 1-(x^-1)/x^1/2 in simplest form without rational denominator (x is an integer)

Jaxxnuts · Jul 29, 2017

1729 · Jul 29, 2017

Jaxxnuts said:
1. For every pair of numbers a&b, the function f(x) satisfies b^2 f(a)=a^2 f(b)
If f(2) is not equal to 0 find
((f(5)-f(1))/2
2. 1-(x^-1)/x^1/2 in simplest form without rational denominator (x is an integer)

$$\noindent \textbf{Q1.} Should be written $\frac{f(5)-f(1)}{f(2)}$, hence why they said $f(2) \neq 0$. Now if $b^2f(a) = a^2f(b)$ then $\frac{f(a)}{f(b)} = \frac{a^2}{b^2}$. \\ \begin{align*} \quad a = 5 \text{ and } b =2 \implies \frac{f(5)}{f(2)} &= \frac{25}{4} \\ a = 1 \text{ and } b = 2 \implies \frac{f(1)}{f(2)} &= \frac{1}{4} \end{align*} \\ Therefore, $\frac{f(5)-f(1)}{f(2)} = \frac{f(5)}{f(2)} - \frac{f(1)}{f(2)} = \frac{25}{4} - \frac{1}{4} = 6.$

$$\noindent \textbf{Q2.} Without fractional/negative indices and with rational denominator, \\ \begin{align*}\quad \frac{1-x^{-1}}{x^{\frac{1}{2}}} &= \frac{1-\frac{1}{x}}{x^{\frac{1}{2}}} \times \frac{x}{x} \\ &= \frac{x-1}{x\sqrt{x}} \\ &= \frac{x-1}{x\sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} \\ &= \frac{x\sqrt{x} - \sqrt{x}}{x^2} \end{align*}$

Jaxxnuts · Jul 30, 2017

Thank you 1729

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