| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Solve using quadratic in trig function |
| Difficulty | Standard +0.3 This is a standard quadratic-in-disguise problem requiring substitution u = cos²x, solving a quadratic equation, and applying domain restrictions. Part (a) involves routine exact value work, while part (b) requires showing discriminant is negative—both are textbook techniques with no novel insight needed. Slightly easier than average due to straightforward structure. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4\cos^4 x + \cos^2 x - 3 = 0 \Rightarrow (4\cos^2 x - 3)(\cos^2 x + 1) = 0\) | M1 | Attempt to solve 3 term quartic (or quadratic in another variable) |
| \(\Rightarrow [\cos^2 x =]\ \frac{3}{4}\ \ [\cos^2 x = -1]\) | A1 | If M0 scored then SC B1 is available for sight of \(\frac{3}{4}\) [and \(-1\)] |
| \(\Rightarrow \cos x = [\pm]\sqrt{\text{their}\ \frac{3}{4}}\) OE \(\left[= \pm\frac{\sqrt{3}}{2}\right]\) | M1 | Square rooting '*their* \(\cos^2 x\)'. Allow without \(\pm\). May be implied by correct final answer(s). Ignore \(\sqrt{-1}\). |
| \([x =]\ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\) | A1 A1 FT | Dependent on preceding M1 only. Exact answers needed. A1 for any 2 correct answers; A1 A1 for 4 correct answers and no others inside range \(0 \leqslant x \leqslant 2\pi\). A1 FT can be awarded for two exact answers that are \(2\pi - \text{'their'}\ \frac{\pi}{6}\) and \(\frac{5\pi}{6}\), within the range \(0 \leqslant x \leqslant 2\pi\). SC: If all 4 answers given in degrees (30, 150, 210, 330) or non-exact (AWRT 0.524, 2.62, 3.67, 5.76 or \(0.167\pi, 0.833\pi, 1.17\pi, 1.83\)) and no others then SC B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos^2 x = \frac{-1-\sqrt{1+16k}}{8} < 0\) [∴ no solutions] | B1 | State that this root is less than 0, needs to be linked to \(\cos^2 x\). Can be achieved by substituting a value for \(k \geqslant 0\). |
| \([\cos^2 x] = \frac{-1 \pm \sqrt{1+16k}}{8}\) | *M1 | Must use quadratic formula. Allow any value of \(k\) but not \(\pm 3\). Condone \(+\) rather than \(\pm\). |
| Substituting \(k = 5\) and obtain 1 from the formula | DM1 | Or argue logically if \(k > 5 \Rightarrow 1 + 16k > 81 \Rightarrow > 1\). |
| \(\cos^2 x = 1\) or \(\cos^2 x > \) or \(\geqslant 1\) | A1 | Needs to be linked to \(\cos^2 x\). |
| Concluding statement having considered both \(\pm\) cases ∴ no solutions | A1 | Dependent upon all previous marks having been scored |
| Alternative: \(\frac{-1+\sqrt{1+16k}}{8} * 1 \Rightarrow -1+\sqrt{1+16k} * 8 \Rightarrow 1+16k * 81\) | DM1 | \(*\) represents any inequality or \(=\) |
| \(k * 5\) | A1 | \(*\) represents any inequality or \(=\) |
| Concluding statement having considered both \(\pm\) cases ∴ no solutions | A1 | Dependent upon all previous marks having been scored |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\cos^4 x + \cos^2 x - 3 = 0 \Rightarrow (4\cos^2 x - 3)(\cos^2 x + 1) = 0$ | M1 | Attempt to solve 3 term quartic (or quadratic in another variable) |
| $\Rightarrow [\cos^2 x =]\ \frac{3}{4}\ \ [\cos^2 x = -1]$ | A1 | If M0 scored then **SC B1** is available for sight of $\frac{3}{4}$ [and $-1$] |
| $\Rightarrow \cos x = [\pm]\sqrt{\text{their}\ \frac{3}{4}}$ **OE** $\left[= \pm\frac{\sqrt{3}}{2}\right]$ | M1 | Square rooting '*their* $\cos^2 x$'. Allow without $\pm$. May be implied by correct final answer(s). Ignore $\sqrt{-1}$. |
| $[x =]\ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ | A1 A1 FT | Dependent on preceding M1 only. Exact answers needed. A1 for any 2 correct answers; A1 A1 for 4 correct answers and no others inside range $0 \leqslant x \leqslant 2\pi$. A1 FT can be awarded for two exact answers that are $2\pi - \text{'their'}\ \frac{\pi}{6}$ and $\frac{5\pi}{6}$, within the range $0 \leqslant x \leqslant 2\pi$. **SC**: If all 4 answers given in degrees (30, 150, 210, 330) or non-exact (AWRT 0.524, 2.62, 3.67, 5.76 or $0.167\pi, 0.833\pi, 1.17\pi, 1.83$) and no others then **SC B1** |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos^2 x = \frac{-1-\sqrt{1+16k}}{8} < 0$ [∴ no solutions] | B1 | State that this root is less than 0, needs to be linked to $\cos^2 x$. Can be achieved by substituting a value for $k \geqslant 0$. |
| $[\cos^2 x] = \frac{-1 \pm \sqrt{1+16k}}{8}$ | *M1 | Must use quadratic formula. Allow any value of $k$ **but not** $\pm 3$. Condone $+$ rather than $\pm$. |
| Substituting $k = 5$ and obtain 1 from the formula | DM1 | Or argue logically if $k > 5 \Rightarrow 1 + 16k > 81 \Rightarrow > 1$. |
| $\cos^2 x = 1$ or $\cos^2 x > $ or $\geqslant 1$ | A1 | Needs to be linked to $\cos^2 x$. |
| Concluding statement having considered both $\pm$ cases ∴ no solutions | A1 | Dependent upon all previous marks having been scored |
| **Alternative:** $\frac{-1+\sqrt{1+16k}}{8} * 1 \Rightarrow -1+\sqrt{1+16k} * 8 \Rightarrow 1+16k * 81$ | DM1 | $*$ represents any inequality or $=$ |
| $k * 5$ | A1 | $*$ represents any inequality or $=$ |
| Concluding statement having considered both $\pm$ cases ∴ no solutions | A1 | Dependent upon all previous marks having been scored |11 The function f is given by $\mathrm { f } ( x ) = 4 \cos ^ { 4 } x + \cos ^ { 2 } x - k$ for $0 \leqslant x \leqslant 2 \pi$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Given that $k = 3$, find the exact solutions of the equation $\mathrm { f } ( x ) = 0$.
\item Use the quadratic formula to show that, when $k > 5$, the equation $\mathrm { f } ( x ) = 0$ has no solutions.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q11 [10]}}