Write down $f'\left( 2 \right)$.

Find $f\left( 2 \right)$.

Show that the graph of g has a gradient of 6 at P.

Let L₂ be the tangent to the graph of g at P. L₁ intersects L₂ at the point Q.

Find the y-coordinate of Q.

recognize that $f'\left( x \right)$ is the gradient of the tangent at $x$     (M1)

eg   $f'\left( x \right) = m$

$f'\left( 2 \right) = 3$  (accept m = 3)     A1 N2

[2 marks]

recognize that $f\left( 2 \right) = y\left( 2 \right)$     (M1)

eg  $f\left( 2 \right) = 3 \times 2 + 1$

$f\left( 2 \right) = 7$     A1 N2

[2 marks]

recognize that the gradient of the graph of g is $g'\left( x \right)$      (M1)

choosing chain rule to find $g'\left( x \right)$      (M1)

eg  $\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u' = 2x$

$g'\left( x \right) = f'\left( {{x^2} + 1} \right) \times 2x$     A2

$g'\left( 1 \right) = 3 \times 2$     A1

$g'\left( 1 \right) = 6$     AG N0

[5 marks]

 at Q, L₁ = L₂ (seen anywhere)      (M1)

recognize that the gradient of L₂ is g'(1)  (seen anywhere)     (M1)
eg  m = 6

finding g (1)  (seen anywhere)      (A1)
eg  $g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7$

attempt to substitute gradient and/or coordinates into equation of a straight line      M1
eg  $y - g\left( 1 \right) = 6\left( {x - 1} \right),\,\,y - 1 = g'\left( 1 \right)\left( {x - 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}$

correct equation for L₂ 

eg  $y - 7 = 6\left( {x - 1} \right),\,\,y = 6x + 1$     A1

correct working to find Q       (A1)
eg   same y-intercept, $3x = 0$

$y = 1$     A1 N2

[7 marks]