All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

$y= \frac{x^{2}+x+1}{x-1}$
$y(x-1) = x^{2}+x+1$
$0 = x^{2}+x-xy+y+1$
For $x\in \mathbb{R}$, y exists only when $b^{2}-4ac \ge 0$

$(1-y)^{2}-4(1)(1+y) \ge 0$

$1-2y+y^{2}-4-4y \ge 0$

$y^{2}-6y-3 \ge 0$

$y \le 3-2\sqrt{3}~\mathrm{or}~y \ge 3+2\sqrt{3}$

### KS Comments:

When I first saw this, I was really surprised as O’levels had something really similar. The above method is the most direct, yes, 5 marks for that. Students can also go and find the turning point (and justify they are maximum or minimum) to decide the range.

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