22. Let $f(x)=x+\sqrt{x^{2}+1}+\frac{1}{x-\sqrt{x^{2}+1}}$.
What is the value of $f(2015)$ ?
A -1
B 0
C 1
D $\sqrt{2016}$
E 2015
Solution
B
Commentary
At first sight, this looks an impossibly difficult question, as it seems to involve working out the value of the square root $\sqrt{x^{2}+1}$ for $x=2015$, without the use of a calculator As this cannot be the intended method, we look for an alternative approach.
The presence of both the terms $x-\sqrt{x^{2}+1}$ and $x+\sqrt{x^{2}+1}$ in the expression for $f(x)$ suggests that we can make progress using algebra, and in particular, the difference of two squares formula $(a-b)(a+b)=a^{2}-b^{2}$, with $a=x$, and $b=\sqrt{x^{2}+1}$.
Indeed, if you have the confidence to try this approach, you will see that this question turns out not to be at all difficult.

We have
$$
f(x)=x+\sqrt{x^{2}+1}+\frac{1}{x-\sqrt{x^{2}+1}}
$$

We now put the two terms involved in $f(x)$ over a common denominator. This gives
$$
\begin{aligned}
f(x) & =\frac{\left(x-\sqrt{x^{2}+1}\right)\left(x+\sqrt{x^{2}+1}\right)+1}{x-\sqrt{x^{2}+1}} \\
& =\frac{\left(x^{2}-\left(\sqrt{x^{2}+1}\right)^{2}\right)+1}{x-\sqrt{x^{2}+1}} \\
& =\frac{\left(x^{2}-\left(x^{2}+1\right)\right)+1}{x-\sqrt{x^{2}+1}} \\
& =\frac{-1+1}{x-\sqrt{x^{2}+1}} \\
& =\frac{0}{x-\sqrt{x^{2}+1}} \\
& =0 .
\end{aligned}
$$

This holds whatever the value of $x$. Therefore, in particular, $f(2015)=0$.