User interface language: English | Español

Date November 2014 Marks available 2 Reference code 14N.2.sl.TZ0.5
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 5 Adapted from N/A

Question

The following diagram shows part of the graph of \(y = p\sin (qx) + r\).

The point \({\text{A}}\left( {\frac{\pi }{6},{\text{ }}2} \right)\) is a maximum point and the point \({\text{B}}\left( {\frac{\pi }{6},{\text{ }}1} \right)\) is a minimum point.

Find the value of

\(p\);

[2]
a.

\(r\);

[2]
b.

\(q\).

[2]
c.

Markscheme

valid approach     (M1)

eg\(\;\;\;\)\(\frac{{2 - 1}}{2},{\text{ }}2 - 1.5\)

\(p = 0.5\)     A1     N2

[2 marks]

a.

valid approach     (M1)

eg\(\;\;\;\)\(\frac{{1 + 2}}{2}\)

\(r = 1.5\)     A1     N2

[2 marks]

b.

METHOD 1

valid approach (seen anywhere)     M1

eg\(\;\;\;\)\(q = \frac{{2\pi }}{{{\text{period}}}},{\text{ }}\frac{{2\pi }}{{\left( {\frac{{2\pi }}{3}} \right)}}\)

period \( = \frac{{2\pi }}{3}\)\(\;\;\;\)(seen anywhere)     (A1)

\(q = 3\)     A1     N2

METHOD 2

attempt to substitute one point and their values for \(p\) and \(r\) into \(y\)     M1

eg\(\;\;\;\)\(2 = 0.5\sin \left( {q\frac{\pi }{6}} \right) + 1.5,{\text{ }}\frac{\pi }{2} = 0.5\sin (q1) + 1.5\)

correct equation in \(q\)     (A1)

eg\(\;\;\;\)\(q\frac{\pi }{6} = \frac{\pi }{2},{\text{ }}q\frac{\pi }{2} = \frac{{3\pi }}{2}\)

\(q = 3\)     A1     N2

METHOD 3

valid reasoning comparing the graph with that of \(\sin x\)     R1

eg\(\;\;\;\)position of max/min, graph goes faster

correct working     (A1)

eg\(\;\;\;\)max at \(\frac{\pi }{6}\) not at \(\frac{\pi }{2}\), graph goes \(3\) times as fast

\(q = 3\)     A1     N2

[3 marks]

Total [7 marks]

c.

Examiners report

Many candidates found the correct value for the amplitude and vertical shift, but very few managed to find the correct value of the period and therefore of \(q\) in part (c). Some candidates substituted the coordinates of a point into the function but were not able to write a correct equation in terms of \(q\). Many candidates who found the correct answer did not show sufficient work to gain all three marks. The rubrics stress the need to show working.

a.

Many candidates found the correct value for the amplitude and vertical shift, but very few managed to find the correct value of the period and therefore of \(q\) in part (c). Some candidates substituted the coordinates of a point into the function but were not able to write a correct equation in terms of \(q\). Many candidates who found the correct answer did not show sufficient work to gain all three marks. The rubrics stress the need to show working.

b.

Many candidates found the correct value for the amplitude and vertical shift, but very few managed to find the correct value of the period and therefore of \(q\) in part (c). Some candidates substituted the coordinates of a point into the function but were not able to write a correct equation in terms of \(q\). Many candidates who found the correct answer did not show sufficient work to gain all three marks. The rubrics stress the need to show working.

c.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.4 » The circular functions \(\sin x\) , \(\cos x\) and \(\tan x\) : their domains and ranges; amplitude, their periodic nature; and their graphs.
Show 59 related questions

View options