Date | May 2018 | Marks available | 5 | Reference code | 18M.1.sl.TZ2.10 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Show that | Question number | 10 | Adapted from | N/A |
Question
Consider a function \(f\). The line L1 with equation \(y = 3x + 1\) is a tangent to the graph of \(f\) when \(x = 2\)
Let \(g\left( x \right) = f\left( {{x^2} + 1} \right)\) and P be the point on the graph of \(g\) where \(x = 1\).
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.
Markscheme
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]