Find the set of values of y for which this equation has real roots.

Hence determine the range of the function \(f:x \to \frac{{x + 1}}{{{x^2} + x + 1}}\).

Explain why f has no inverse.

for the equation to have real roots

\({(y - 1)^2} - 4y(y - 1) \geqslant 0\)     M1

\( \Rightarrow 3{y^2} - 2y - 1 \leqslant 0\)

(by sign diagram, or algebraic method)     M1

\( - \frac{1}{3} \leqslant y \leqslant 1\)     A1A1

Note: Award first A1 for \( - \frac{1}{3}\) and 1, and second A1 for inequalities. These are independent marks.

[4 marks]

\(f:x \to \frac{{x + 1}}{{{x^2} + x + 1}} \Rightarrow x + 1 = y{x^2} + yx + y\)     (M1)

\( \Rightarrow 0 = y{x^2} + (y - 1)x + (y - 1)\)     A1

hence, from (a) range is \( - \frac{1}{3} \leqslant y \leqslant 1\)     A1

[3 marks]

a value for y would lead to 2 values for x from (a)     R1

Note: Do not award R1 if (b) has not been tackled.

[1 mark]

(a) The best answered part of the question. The critical points were usually found, but the inequalities were often incorrect. Few candidates were convincing regarding the connection between (a) and (b). This had consequences for (c).

(a) The best answered part of the question. The critical points were usually found, but the inequalities were often incorrect. Few candidates were convincing regarding the connection between (a) and (b). This had consequences for (c).

(a) The best answered part of the question. The critical points were usually found, but the inequalities were often incorrect. Few candidates were convincing regarding the connection between (a) and (b). This had consequences for (c).