Easy -1.2 This is a straightforward single-step differentiation problem requiring the chain rule to find dy/dx, evaluate at x=0, find the y-coordinate, then write the tangent equation using y-y₁=m(x-x₁). It's simpler than average A-level questions as it involves only basic differentiation and no problem-solving insight.
Correct derivative; PI by \(-32\) obtained with no errors in evaluating \(\frac{dy}{dx}\)
When \(x = 0\), \(\frac{dy}{dx} = -32\); \(y = 16\)
M1
Substitutes \(x = 0\) into \(\frac{dy}{dx}\) to obtain numerical value; or PI by constant from \(\frac{dy}{dx}\); or PI by \(-32\) with no errors
\(y = -32x + 16\)
A1
ACF; award at first opportunity; ISW incorrect rearrangement; no errors seen
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 4(x-2)^3$ or $4x^3 - 24x^2 + 48x - 32$ | B1 | Correct derivative; PI by $-32$ obtained with no errors in evaluating $\frac{dy}{dx}$ |
| When $x = 0$, $\frac{dy}{dx} = -32$; $y = 16$ | M1 | Substitutes $x = 0$ into $\frac{dy}{dx}$ to obtain numerical value; or PI by constant from $\frac{dy}{dx}$; or PI by $-32$ with no errors |
| $y = -32x + 16$ | A1 | ACF; award at first opportunity; ISW incorrect rearrangement; no errors seen |
---