| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve shifted trig equation |
| Difficulty | Moderate -0.8 This C2 question tests basic transformations (stretch, reflection, translation), a routine shifted trig equation requiring one inverse sine calculation, and a straightforward algebraic identity proof using expansion. All parts are standard textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{12 + x^2\sqrt{x}}{x} = 12x^{-1} + x^{\frac{3}{2}}\) | B1, B1, B1 | \(p=-1\), \(q=\frac{3}{2}\) |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = -12x^{-2} + \frac{3}{2}x^{\frac{1}{2}}\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(x=4\): \(\frac{dy}{dx} = -12(16)^{-1} + \frac{3}{2}(2) = -\frac{3}{4} + 3 = \frac{9}{4}\) | M1, A1 | |
| Gradient of normal \(= -\frac{4}{9}\) | M1 | |
| \(y\) at \(x=4\): \(y = \frac{12+4^2\cdot 2}{4} = \frac{12+32}{4} = 11\) | ||
| Normal: \(y - 11 = -\frac{4}{9}(x-4)\) | A1 | oe |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 0 \Rightarrow -12x^{-2} + \frac{3}{2}x^{\frac{1}{2}} = 0\) | M1 |
| \(\frac{3}{2}x^{\frac{1}{2}} = \frac{12}{x^2} \Rightarrow x^{\frac{5}{2}} = 8\) | A1 |
| \(x = 8^{\frac{2}{5}} = 2^{\frac{6}{5}} = 2^k\), \(k = \frac{6}{5}\) | A1 |
# Question 6:
**(a)**
$\frac{12 + x^2\sqrt{x}}{x} = 12x^{-1} + x^{\frac{3}{2}}$ | B1, B1, B1 | $p=-1$, $q=\frac{3}{2}$
**(b)(i)**
$\frac{dy}{dx} = -12x^{-2} + \frac{3}{2}x^{\frac{1}{2}}$ | M1, A1 |
**(b)(ii)**
At $x=4$: $\frac{dy}{dx} = -12(16)^{-1} + \frac{3}{2}(2) = -\frac{3}{4} + 3 = \frac{9}{4}$ | M1, A1 |
Gradient of normal $= -\frac{4}{9}$ | M1 |
$y$ at $x=4$: $y = \frac{12+4^2\cdot 2}{4} = \frac{12+32}{4} = 11$ | |
Normal: $y - 11 = -\frac{4}{9}(x-4)$ | A1 | oe
**(b)(iii)**
$\frac{dy}{dx} = 0 \Rightarrow -12x^{-2} + \frac{3}{2}x^{\frac{1}{2}} = 0$ | M1 |
$\frac{3}{2}x^{\frac{1}{2}} = \frac{12}{x^2} \Rightarrow x^{\frac{5}{2}} = 8$ | A1 |
$x = 8^{\frac{2}{5}} = 2^{\frac{6}{5}} = 2^k$, $k = \frac{6}{5}$ | A1 |
---6
\begin{enumerate}[label=(\alph*)]
\item Describe the geometrical transformation that maps the curve with equation $y = \sin x$ onto the curve with equation:
\begin{enumerate}[label=(\roman*)]
\item $y = 2 \sin x$;
\item $y = - \sin x$;
\item $\quad y = \sin \left( x - 30 ^ { \circ } \right)$.
\end{enumerate}\item Solve the equation $\sin \left( \theta - 30 ^ { \circ } \right) = 0.7$, giving your answers to the nearest $0.1 ^ { \circ }$ in the interval $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\item Prove that $( \cos x + \sin x ) ^ { 2 } + ( \cos x - \sin x ) ^ { 2 } = 2$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 Q6}}