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 $\frac{\pi }{2}$.