The normal to the curve \(x{{\text{e}}^{ - y}} + {{\text{e}}^y} = 1 + x\), at the point (c, \(\ln c\)), has a y-intercept \({c^2} + 1\).

Determine the value of c.

EITHER

differentiating implicitly:

\(1 \times {{\text{e}}^{ - y}} - x{{\text{e}}^{ - y}}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {{\text{e}}^y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1\)     M1A1

at the point (c, \(\ln c\))

\(\frac{1}{c} - c \times \frac{1}{c}\frac{{{\text{d}}y}}{{{\text{d}}x}} + c\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{c}\,\,\,\,\,(c \ne 1)\)     (A1)

OR

reasonable attempt to make expression explicit     (M1)

\(x{{\text{e}}^{ - y}} + {{\text{e}}^y} = 1 + x\)

\(x + {{\text{e}}^{2y}} = {{\text{e}}^y}(1 + x)\)

\({{\text{e}}^{2y}} - {{\text{e}}^y}(1 + x) + x = 0\)

\(({{\text{e}}^y} - 1)({{\text{e}}^y} - x) = 0\)     (A1)

\({{\text{e}}^y} = 1,{\text{ }}{{\text{e}}^y} = x\)

\(y = 0,{\text{ }}y = \ln x\)     A1

Note: Do not penalize if y = 0 not stated.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{x}\)

gradient of tangent \( = \frac{1}{c}\)     A1

Note: If candidate starts with \(y = \ln x\) with no justification, award (M0)(A0)A1A1.

THEN

the equation of the normal is

\(y - \ln c = - c(x - c)\)     M1

\(x = 0,{\text{ }}y = {c^2} + 1\)

\({c^2} + 1 - \ln c = {c^2}\)     (A1)

\(\ln c = 1\)

\(c = {\text{e}}\)     A1

[7 marks]

This was the first question to cause the majority of candidates a problem and only the better candidates gained full marks. Weaker candidates made errors in the implicit differentiation and those who were able to do this often were unable to simplify the expression they gained for the gradient of the normal in terms of c; a significant number of candidates did not know how to simplify the logarithms appropriately.