Date | May 2019 | Marks available | 4 | Reference code | 19M.1.SL.TZ1.S_10 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Find | Question number | S_10 | Adapted from | N/A |
Question
Consider f(x)=√xsin(π4x) and g(x)=√x for x ≥ 0. The first time the graphs of f and g intersect is at x=0.
The set of all non-zero values that satisfy f(x)=g(x) can be described as an arithmetic sequence, un=a+bn where n ≥ 1.
Find the two smallest non-zero values of x for which f(x)=g(x).
At point P, the graphs of f and g intersect for the 21st time. Find the coordinates of P.
The following diagram shows part of the graph of g reflected in the x-axis. It also shows part of the graph of f and the point P.
Find an expression for the area of the shaded region. Do not calculate the value of the expression.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct working (A1)
eg sin(π4x)=1, √x(1−sin(π4x))=0
sin(π2)=1 (seen anywhere) (A1)
correct working (ignore additional values) (A1)
eg π4x=π2, π4x=π2+2π
x = 2, 10 A1A1 N1N1
[5 marks]
valid approach (M1)
eg first intersection at x=0, n=20
correct working A1
eg −6+8×20, 2+(20−1)×8, u20=154
P(154, √154) (accept x=154 and y=√154) A1A1 N3
[4 marks]
valid attempt to find upper boundary (M1)
eg half way between u20 and u21, u20+d2, 154 + 4, −2+8n, at least two values of new sequence {6, 14, ...}
upper boundary at x=158 (seen anywhere) (A1)
correct integral expression (accept missing dx) A1A1 N4
eg ∫1580(√xsin(π4x)+√x)dx, ∫1580(g+f)dx), ∫1580√xsin(π4x)dx−∫1580−√xdx
Note: Award A1 for two correct limits and A1 for correct integrand. The A1 for correct integrand may be awarded independently of all the other marks.
[4 marks]