Date | May 2021 | Marks available | 3 | Reference code | 21M.1.SL.TZ1.7 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Hence and Find | Question number | 7 | Adapted from | N/A |
Question
Let f(x)=mx2-2mx, where x∈ℝ and m∈ℝ. The line y=mx-9 meets the graph of f at exactly one point.
The function f can be expressed in the form f(x)=4(x-p)(x-q), where p, q∈ℝ.
The function f can also be expressed in the form f(x)=4(x-h)2+k, where h, k∈ℝ.
Show that m=4.
Find the value of p and the value of q.
Find the value of h and the value of k.
Hence find the values of x where the graph of f is both negative and increasing.
Markscheme
METHOD 1 (discriminant)
mx2-2mx=mx-9 (M1)
mx2-3mx+9=0
recognizing Δ=0 (seen anywhere) M1
Δ=(-3m)2-4(m)(9) (do not accept only in quadratic formula for x) A1
valid approach to solve quadratic for m (M1)
9m(m-4)=0 OR m=36±√362-4×9×02×9
both solutions m=0, 4 A1
m≠0 with a valid reason R1
the two graphs would not intersect OR 0≠-9
m=4 AG
METHOD 2 (equating slopes)
mx2-2mx=mx-9 (seen anywhere) (M1)
f'(x)=2mx-2m A1
equating slopes, f'(x)=m (seen anywhere) M1
2mx-2m=m
x=32 A1
substituting their x value (M1)
(32)2m-2m×32=m×32-9
94m-124m=64m-9 A1
-9m4=-9
m=4 AG
METHOD 3 (using -b2a)
mx2-2mx=mx-9 (M1)
mx2-3mx+9=0
attempt to find x-coord of vertex using -b2a (M1)
-(-3m)2m A1
x=32 A1
substituting their x value (M1)
(32)2m-3m×32+9=0
94m-92m+9=0 A1
-9m=-36
m=4 AG
[6 marks]
4x(x-2) (A1)
p=0 and q=2 OR p=2 and q=0 A1
[2 marks]
attempt to use valid approach (M1)
0+22, OR
A1A1
[3 marks]
EITHER
recognition to (may be seen on sketch) (M1)
OR
recognition that and (M1)
THEN
A1A1
Note: Award A1 for two correct values, A1 for correct inequality signs.
[3 marks]