Date | May 2012 | Marks available | 4 | Reference code | 12M.1.sl.TZ2.13 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 13 | Adapted from | N/A |
Question
The equation of a curve is given as \(y = 2x^{2} - 5x + 4\).
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).
The equation of the line L is \(6x + 2y = -1\).
Find the x-coordinate of the point on the curve \(y = 2x^2 - 5x + 4\) where the tangent is parallel to L.
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4x - 5\) (A1)(A1) (C2)
Notes: Award (A1) for each correct term. Award (A1)(A0) if any other terms are given.
[2 marks]
\(y = - 3x - \frac{1}{2}\) (M1)
Note: Award (M1) for rearrangement of equation
gradient of line is –3 (A1)
\(4x - 5 = -3\) (M1)
Notes: Award (M1) for equating their gradient to their derivative from part (a). If \(4x - 5 = -3\) is seen with no working award (M1)(A1)(M1).
\(x = \frac{1}{2}\) (A1)(ft) (C4)
Note: Follow through from their part (a). If answer is given as (0.5, 2) with no working award the final (A1) only.
[4 marks]
Examiners report
The derivative of the function was correctly found by most candidates. Rearranging the equation of the line to find the gradient was also successfully performed. Most candidates could not find the x-coordinate of the point on the curve whose tangent was parallel to a given line. To most candidates, part (b) appeared to be disconnected to part (a).
The derivative of the function was correctly found by most candidates. Rearranging the equation of the line to find the gradient was also successfully performed. Most candidates could not find the x-coordinate of the point on the curve whose tangent was parallel to a given line. To most candidates, part (b) appeared to be disconnected to part (a).