CAIE P1 2002 June — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2002
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Graphs & Exact Values
TypeFind function constants from given conditions
DifficultyEasy -1.2 This is a straightforward question requiring substitution of exact trig values (sin(π/2)=1, sin(3π/2)=-1) to form two simultaneous equations, then solving for a and b. Part (ii) involves basic equation solving and calculator work, while part (iii) is a routine sketch. All steps are standard procedures with no problem-solving insight required, making it easier than average.
Spec1.02m Graphs of functions: difference between plotting and sketching1.02u Functions: definition and vocabulary (domain, range, mapping)1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals

6 The function f , where \(\mathrm { f } ( x ) = a \sin x + b\), is defined for the domain \(0 \leqslant x \leqslant 2 \pi\). Given that \(\mathrm { f } \left( \frac { 1 } { 2 } \pi \right) = 2\) and that \(\mathrm { f } \left( \frac { 3 } { 2 } \pi \right) = - 8\),
  1. find the values of \(a\) and \(b\),
  2. find the values of \(x\) for which \(\mathrm { f } ( x ) = 0\), giving your answers in radians correct to 2 decimal places,
  3. sketch the graph of \(y = \mathrm { f } ( x )\).

\(f(x) = a\sin x + b\)
(i) \(f(\pi/2) = 2\)
\(f(3\pi/2) = -8\)
AnswerMarks Guidance
Solution: \(a = -5\), \(b = -3\)B1, B1, B1 Correct only; Correct only; Correct only
(ii) \(5\sin x - 3 = 0\), \(\sin x = \frac{3}{5}\)
AnswerMarks Guidance
\(x = 0.64\) or \(x = 2.50\)B1\*, B1\* For \(\sin^{-1}(-b/a)\); For \(\pi\) - his answer
(iii) GraphB2\*, 2 Just one cycle. Starts on negative y-axis. Max about correct. Min about correct
$f(x) = a\sin x + b$

**(i)** $f(\pi/2) = 2$
$f(3\pi/2) = -8$
Solution: $a = -5$, $b = -3$ | B1, B1, B1 | Correct only; Correct only; Correct only

**(ii)** $5\sin x - 3 = 0$, $\sin x = \frac{3}{5}$
$x = 0.64$ or $x = 2.50$ | B1\*, B1\* | For $\sin^{-1}(-b/a)$; For $\pi$ - his answer

**(iii)** Graph | B2\*, 2 | Just one cycle. Starts on negative y-axis. Max about correct. Min about correct
6 The function f , where $\mathrm { f } ( x ) = a \sin x + b$, is defined for the domain $0 \leqslant x \leqslant 2 \pi$. Given that $\mathrm { f } \left( \frac { 1 } { 2 } \pi \right) = 2$ and that $\mathrm { f } \left( \frac { 3 } { 2 } \pi \right) = - 8$,\\
(i) find the values of $a$ and $b$,\\
(ii) find the values of $x$ for which $\mathrm { f } ( x ) = 0$, giving your answers in radians correct to 2 decimal places,\\
(iii) sketch the graph of $y = \mathrm { f } ( x )$.

\hfill \mbox{\textit{CAIE P1 2002 Q6 [7]}}