CAIE P1 2004 November — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2004
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeShow then solve: sin²/cos² substitution
DifficultyStandard +0.3 This is a straightforward multi-part question requiring standard techniques: using the identity cos²x = 1 - sin²x to simplify, solving a quadratic equation in sin x, and finding range from the simplified form. All steps are routine for P1 level with no novel insight required, making it slightly easier than average.
Spec1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

6 The function \(\mathrm { f } : x \mapsto 5 \sin ^ { 2 } x + 3 \cos ^ { 2 } x\) is defined for the domain \(0 \leqslant x \leqslant \pi\).
  1. Express \(\mathrm { f } ( x )\) in the form \(a + b \sin ^ { 2 } x\), stating the values of \(a\) and \(b\).
  2. Hence find the values of \(x\) for which \(\mathrm { f } ( x ) = 7 \sin x\).
  3. State the range of f .

Question 6:
Part (i)
AnswerMarks Guidance
\(5s^2 + 3c^2 = 5s^2 + 3(1-s^2) \rightarrow 3 + 2\sin^2 x\), \(a=3\), \(b=2\)M1, A1 [2] Use of \(s^2 + c^2 = 1\); \(3 + 2\sin^2 x\) gets both marks
Part (ii)
AnswerMarks Guidance
\(3 + 2s^2 = 7s\); sets to 0 and solves; \(s = \frac{1}{2}\) or \(s = 3\)M1 Sets to \(0\) + correct method of solution
Only values are \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\)A1A1\(\sqrt{}\) [3] Co for one value; other \(\pi = \) "1st"; if degrees give A0,A1\(\sqrt{}\) for \(180-\)
Part (iii)
AnswerMarks Guidance
Minimum value \(= "a" = 3\); Maximum value \(= "a+b" = 5\)
Range \(3 \leq f(x) \leq 5\)B1\(\sqrt{}\)B1\(\sqrt{}\) [2] For his "\(a\)" and "\(a+b\)"; condone \(<\); allow 3 and 5 on their own
# Question 6:

## Part (i)
| $5s^2 + 3c^2 = 5s^2 + 3(1-s^2) \rightarrow 3 + 2\sin^2 x$, $a=3$, $b=2$ | M1, A1 [2] | Use of $s^2 + c^2 = 1$; $3 + 2\sin^2 x$ gets both marks |

## Part (ii)
| $3 + 2s^2 = 7s$; sets to 0 and solves; $s = \frac{1}{2}$ or $s = 3$ | M1 | Sets to $0$ + correct method of solution |
| Only values are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ | A1A1$\sqrt{}$ [3] | Co for one value; other $\pi = $ "1st"; if degrees give A0,A1$\sqrt{}$ for $180-$ |

## Part (iii)
| Minimum value $= "a" = 3$; Maximum value $= "a+b" = 5$ | | |
| Range $3 \leq f(x) \leq 5$ | B1$\sqrt{}$B1$\sqrt{}$ [2] | For his "$a$" and "$a+b$"; condone $<$; allow 3 and 5 on their own |

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6 The function $\mathrm { f } : x \mapsto 5 \sin ^ { 2 } x + 3 \cos ^ { 2 } x$ is defined for the domain $0 \leqslant x \leqslant \pi$.\\
(i) Express $\mathrm { f } ( x )$ in the form $a + b \sin ^ { 2 } x$, stating the values of $a$ and $b$.\\
(ii) Hence find the values of $x$ for which $\mathrm { f } ( x ) = 7 \sin x$.\\
(iii) State the range of f .

\hfill \mbox{\textit{CAIE P1 2004 Q6 [7]}}