The line \(y = m(x - m)\) is a tangent to the curve \((1 - x)y = 1\).

Determine m and the coordinates of the point where the tangent meets the curve.

EITHER

\(y = \frac{1}{{1 - x}} \Rightarrow y' = \frac{1}{{{{(1 - x)}^2}}}\)     M1A1

solve simultaneously     M1

\(\frac{1}{{1 - x}} = m(x - m){\text{ and }}\frac{1}{{{{(1 - x)}^2}}} = m\)

\(\frac{1}{{1 - x}} = \frac{1}{{{{(1 - x)}^2}}}\left( {x - \frac{1}{{{{(1 - x)}^2}}}} \right)\)     A1

Note: Accept equivalent forms.

\({(1 - x)^3} - x{(1 - x)^2} + 1 = 0,{\text{ }}x \ne 1\)

\(x = 1.65729 \ldots  \Rightarrow y = \frac{1}{{1 - 1.65729 \ldots }} = - 1.521379 \ldots \)

tangency point (1.66, –1.52)     A1A1

\(m = {( - 1.52137 \ldots )^2} = 2.31\)     A1

OR

\((1 - x)y = 1\)

\(m(1 - x)(x - m) = 1\)     M1

\(m(x - {x^2} - m + mx) = 1\)

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

\({b^2} - 4ac = 0\)     (M1)

\({(m + {m^2})^2} - 4m({m^2} + 1) = 0\)

\(m = 2.31\)     A1

substituting \(m = 2.31 \ldots {\text{ into }}m{x^2} - x(m + {m^2}) + ({m^2} + 1) = 0\)     (M1)

\(x = 1.66\)     A1

\(y = \frac{1}{{1 - 1.65729}} = - 1.52\)     A1

tangency point (1.66, –1.52)

[7 marks]

Very few candidates answered this question well but among those a variety of nice approaches were seen. This question required some organized thinking and good understanding of the concepts involved and therefore just strong candidates were able to go beyond the first steps. Sadly a few good answers were spoiled due to early rounding.