Date | May 2022 | Marks available | 2 | Reference code | 22M.1.SL.TZ1.6 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Describe | Question number | 6 | Adapted from | N/A |
Question
Consider f(x)=4 sin x+2.5 and g(x)=4 sin(x-3π2)+2.5+q, where x∈ℝ and q>0.
The graph of g is obtained by two transformations of the graph of f.
Describe these two transformations.
The y-intercept of the graph of g is at (0, r).
Given that g(x)≥7, find the smallest value of r.
Markscheme
translation (shift) by 3π2 to the right/positive horizontal direction A1
translation (shift) by q upwards/positive vertical direction A1
Note: accept translation by (3π2q)
Do not accept ‘move’ for translation/shift.
[2 marks]
METHOD 1
minimum of 4 sin(x-3π2) is -4 (may be seen in sketch) (M1)
-4+2.5+q≥7
q≥8.5 (accept q=8.5) A1
substituting x=0 and their q (=8.5) to find r (M1)
(r=) 4 sin(-3π2)+2.5+8.5
4+2.5+8.5 (A1)
smallest value of r is 15 A1
METHOD 2
substituting x=0 to find an expression (for r) in terms of q (M1)
(g(0)=r=) 4 sin(-3π2)+2.5+q
(r=) 6.5+q A1
minimum of 4 sin(x-3π2) is -4 (M1)
-4+2.5+q≥7
-4+2.5+(r-6.5)≥7 (accept =) (A1)
smallest value of r is 15 A1
METHOD 3
4 sin(x-3π2)+2.5+q=4 cos x+2.5+q A1
y-intercept of 4 cos x+2.5+q is a maximum (M1)
amplitude of g(x) is 4 (A1)
attempt to find least maximum (M1)
r=2×4+7
smallest value of r is 15 A1
[5 marks]
Examiners report
Candidates knew aspects of the transformations performed but some were unable to correctly describe them fully, e.g., omitting direction (right/up/positive) or using 'move' instead of translate/shift. Each description requires three parts: transformation type, size and direction. e.g., translation of q units up. For part (b) few candidates were able to fully navigate the reasoning required in this question. A common error was to evaluate sin(-3π2)=-1, instead of 1. Those who used sketches to assist in their thinking were typically more successful.