Review question

# Can we prove these bounds for this fraction involving surds? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7077

## Solution

1. The variables $x$ and $y$ are connected by the relation $y = ax^n$, where $a$ and $n$ are constants; $y = 3$ when $x = 4$ and $y = 2$ when $x = 9$. Find the exact values of $n$ and $a$.
By substituting the information we’re given in the question, we obtain the equations $$$3 = 4^n a \qquad \label{eq:1}$$$ and $$$2 = 9^n a. \qquad \label{eq:2}$$$

Dividing the second by the first to eliminate $a$ gives $\frac{2}{3} = \left(\frac{9}{4}\right)^n.$

To work with this, it will be more convenient to write everything as powers of $2$ and $3$.

That is, $2^{2n+1}=3^{2n+1}$, which can only be the case when $2n+1=0$ (as $2$ and $3$ have highest common factor $1$).

Therefore, $n=-\frac{1}{2}$.

Now we can go back to equation $\eqref{eq:1}$ or equation $\eqref{eq:2}$ to find that $a=6$.

Alternatively, if you know about logarithms…

Taking logs to base $10$ with our starting equations (any base will do, in fact) we get $\log3 = \log a + 2n\log2, \log2 = \log a + 2n \log3.$

Subtracting these gives $\log3 - \log2 = 2n(\log2-\log3)$, and so $n = -\dfrac{1}{2}$.

Substituting back into one of our starting equations gives $a = 6$.

1. Express $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$ as a fraction having a rational denominator.

If we multiply both the numerator and the denominator by $\sqrt{a} + \sqrt{b}$, then the fraction becomes $\dfrac{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{(\sqrt{a}+\sqrt{b})^2}{a-b}=\dfrac{a+b+2\sqrt{ab}}{a-b}.$ This has a rational denominator.

1. Show that, if $n$ is a positive integer, then $n(n+2)$ lies between $n^2$ and $(n+1)^2$.

Note $n(n+2)-n^2 = 2n > 0$, so $n(n+2)>n^2$.

We also have $n(n+2)-(n+1)^2=n^2 + 2n - n^2 - 2n -1=-1 < 0,$ so $n(n+2)<(n+1)^2$.

So $n(n+2)$ lies between $n^2$ and $(n+1)^2$.

1. If $n$ is a positive integer, use the results in (i) and (ii) to find in terms of $n$ two consecutive integers between which $\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$ must lie.

By using (i) we get $\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}=\dfrac{2n+2+2\sqrt{n(n+2)}}{2}=n+1+\sqrt{(n+2)n}.$

Now we know $(n+1)^2>n(n+2)>n^2$, which implies $n+1>\sqrt{n(n+2)}>n$.

So adding $n+1$ to each part of the inequality, we get $2n+2>n+1+\sqrt{(n+2)n}>2n+1,$

and so $\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}$ lies between $2n+1$ and $2n+2$.