Date | November 2019 | Marks available | 7 | Reference code | 19N.1.SL.TZ0.S_10 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | S_10 | Adapted from | N/A |
Question
Let g(x)=px+q, for x, p, q∈R, p>1. The point A(0, a) lies on the graph of g.
Let f(x)=g−1(x). The point B lies on the graph of f and is the reflection of point A in the line y=x.
The line L1 is tangent to the graph of f at B.
Write down the coordinates of B.
Given that f′(a)=1lnp, find the equation of L1 in terms of x, p and q.
The line L2 is tangent to the graph of g at A and has equation y=(lnp)x+q+1.
The line L2 passes through the point (−2, −2).
The gradient of the normal to g at A is 1ln(13).
Find the equation of L1 in terms of x.
Markscheme
B(a, 0) (accept B(q+1, 0)) A2 N2
[2 marks]
Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding a.
FINDING a
valid attempt to find an expression for a in terms of q (M1)
g(0)=a, p0+q=a
a=q+1 (A1)
FINDING THE EQUATION OF L1
EITHER
attempt to substitute tangent gradient and coordinates into equation of straight line (M1)
eg y−0=f′(a)(x−a), y=f′(a)(x−(q+1))
correct equation in terms of a and p (A1)
eg y−0=1ln(p)(x−a)
OR
attempt to substitute tangent gradient and coordinates to find b
eg 0=1ln(p)(a)+b
b=−aln(p) (A1)
THEN (must be in terms of both p and q)
y=1lnp(x−q−1), y=1lnpx−q+1lnp A1 N3
Note: Award A0 for final answers in the form L1=1lnp(x−q−1)
[5 marks]
Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find q in terms of p before finding a value for p.
FINDING p
valid approach to find the gradient of the tangent (M1)
eg m1m2=−1, −11ln(13), −ln(13), −1lnp=1ln(13)
correct application of log rule (seen anywhere) (A1)
eg ln(13)−1, −(ln(1)−ln(3))
correct equation (seen anywhere) A1
eg lnp=ln3, p=3
FINDING q
correct substitution of (−2, −2) into L2 equation (A1)
eg −2=(lnp)(−2)+q+1
q=2lnp−3, q=2ln3−3 (seen anywhere) A1
FINDING L1
correct substitution of their p and q into their L1 (A1)
eg y=1ln3(x−(2ln3−3)−1)
y=1ln3(x−2ln3+2), y=1ln3x−2ln3−2ln3 A1 N2
Note: Award A0 for final answers in the form L1=1ln3(x−2ln3+2).
[7 marks]