A normal to the graph of \(y = \arctan (x - 1)\) , for \(x > 0\), has equation \(y = - 2x + c\) , where \(x \in \mathbb{R}\) .

Find the value of c.

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\arctan (x - 1)} \right) = \frac{1}{{1 + {{(x - 1)}^2}}}\)   (or equivalent)     A1

\({m_N} = - 2{\text{ and so }}{m_T} = \frac{1}{2}\)     (R1)

Attempting to solve \(\frac{1}{{1 + {{(x - 1)}^2}}} = \frac{1}{2}\) (or equivalent) for x     M1

\(x = 2{\text{ (as }}x > 0)\)     A1

Substituting \(x = 2\) and \(y = \frac{\pi }{4}\) to find c     M1

\(c = 4 + \frac{\pi }{4}\)     A1     N1

[6 marks]

There was a disappointing response to this question from a fair number of candidates. The differentiation was generally correctly performed, but it was then often equated to \( - 2x + c\) rather than the correct numerical value. A few candidates either didn’t simplify arctan(1) to \(\frac{\pi }{4}\), or stated it to be 45 or \(\frac{\pi }{2}\).