Date | November 2015 | Marks available | 2 | Reference code | 15N.1.hl.TZ0.4 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Consider the curve \(y = \frac{1}{{1 - x}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne 1\).
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).
Determine the equation of the normal to the curve at the point \(x = 3\) in the form \(ax + by + c = 0\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}\).
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(1 - x)^{ - 2}}\;\;\;\left( { = \frac{1}{{{{(1 - x)}^2}}}} \right)\) (M1)A1
[2 marks]
gradient of Tangent \( = \frac{1}{4}\) (A1)
gradient of Normal \( = - 4\) (M1)
\(y + \frac{1}{2} = - 4(x - 3)\) or attempt to find \(c\) in \(y = mx + c\) M1
\(8x + 2y - 23 = 0\) A1
[4 marks]
Total [6 marks]