# Question 7:

## Part (a)(i): Find an expression for $\frac{dy}{dx}$

| Working/Answer | Mark | Guidance |
|---|---|---|
| Differentiating implicitly: $-2\sin 2y \cdot \frac{dy}{dx} + e^{3x}\frac{dy}{dx} + 3ye^{3x} = 0$ | M1 | Attempt implicit differentiation of at least one term |
| $-2\sin 2y \cdot \frac{dy}{dx}$ correct | A1 | |
| $e^{3x}\frac{dy}{dx} + 3ye^{3x}$ correct (product rule applied) | M1A1 | M1 for product rule attempt, A1 correct |
| $\frac{dy}{dx}(-2\sin 2y + e^{3x}) = -3ye^{3x}$ | M1 | Collecting $\frac{dy}{dx}$ terms |
| $\frac{dy}{dx} = \frac{-3ye^{3x}}{e^{3x} - 2\sin 2y}$ | A1 | Correct expression |

## Part (a)(ii): Exact value of gradient at $A$

| Working/Answer | Mark | Guidance |
|---|---|---|
| Substituting $x = \ln 2$, $y = \frac{\pi}{4}$: $e^{3\ln 2} = 8$, $\sin\frac{\pi}{2} = 1$ | | |
| $\frac{dy}{dx} = \frac{-3 \cdot \frac{\pi}{4} \cdot 8}{8 - 2} = \frac{-6\pi}{6} = -\pi$ | A1 | ft from (a)(i) |

## Part (b): $y$-coordinate of $B$

| Working/Answer | Mark | Guidance |
|---|---|---|
| Gradient of normal $= \frac{1}{\pi}$ | M1 | Use of $m_1 m_2 = -1$ with their gradient |
| Equation of normal: $y - \frac{\pi}{4} = \frac{1}{\pi}(x - \ln 2)$ | | |
| At $x = 0$: $y = \frac{\pi}{4} - \frac{\ln 2}{\pi}$ | A1 | Exact value required |

I can see these are blank answer pages (pages 22-24) from an AQA MPC4 June 2014 exam paper - they contain no mark scheme content, only empty answer lines for students to write in.

To find the mark scheme for this paper, I'd suggest checking:
- **AQA's website** (aqa.org.uk) under past papers
- **Physics & Maths Tutor** (physicsandmathstutor.com)
- **The Student Room** resources

Would you like me to instead **work through the solutions** to Questions 8(a) and 8(b) shown on page 22, presenting them in a mark-scheme style format?