| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Easy -1.3 This is a straightforward C1 differentiation question requiring only routine application of the power rule for polynomials, evaluation at given points, and standard interpretation of derivatives. All parts are textbook exercises with no problem-solving or novel insight required, making it easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 10x^4 + 20x^3\) | B1 B1 | One mark per term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = 40x^3 + 60x^2\) | B1 | Follow through from (a)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(x=-1\): \(\frac{dy}{dx} = 10(1) + 20(-1) = -10\) | M1 | Substituting \(x=-1\) into \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} < 0\) so \(y\) is decreasing at \(P\) | A1 | Must state decreasing with reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y\)-coordinate at \(P\): \(y = 2(-1) + 5(1) - 1 = 2\) | B1 | |
| Tangent: \(y - 2 = -10(x+1)\) | M1 | Using gradient from (b)(i) |
| \(y = -10x - 8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(x=-2\): \(\frac{dy}{dx} = 10(16) + 20(-8) = 160 - 160 = 0\) | M1 A1 | Show gradient is zero |
| \(\frac{d^2y}{dx^2} = 40(-8) + 60(4) = -320 + 240 = -80\) | M1 | Evaluate second derivative |
| \(\frac{d^2y}{dx^2} < 0\) therefore \(Q\) is a maximum | A1 | Conclusion required |
# Question 3:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 10x^4 + 20x^3$ | B1 B1 | One mark per term |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2y}{dx^2} = 40x^3 + 60x^2$ | B1 | Follow through from (a)(i) |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $x=-1$: $\frac{dy}{dx} = 10(1) + 20(-1) = -10$ | M1 | Substituting $x=-1$ into $\frac{dy}{dx}$ |
| $\frac{dy}{dx} < 0$ so $y$ is decreasing at $P$ | A1 | Must state decreasing with reason |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y$-coordinate at $P$: $y = 2(-1) + 5(1) - 1 = 2$ | B1 | |
| Tangent: $y - 2 = -10(x+1)$ | M1 | Using gradient from (b)(i) |
| $y = -10x - 8$ | A1 | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| At $x=-2$: $\frac{dy}{dx} = 10(16) + 20(-8) = 160 - 160 = 0$ | M1 A1 | Show gradient is zero |
| $\frac{d^2y}{dx^2} = 40(-8) + 60(4) = -320 + 240 = -80$ | M1 | Evaluate second derivative |
| $\frac{d^2y}{dx^2} < 0$ therefore $Q$ is a maximum | A1 | Conclusion required |
---3 A curve has equation $y = 2 x ^ { 5 } + 5 x ^ { 4 } - 1$.
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} x }$
\item $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$
\end{enumerate}\item The point on the curve where $x = - 1$ is $P$.
\begin{enumerate}[label=(\roman*)]
\item Determine whether $y$ is increasing or decreasing at $P$, giving a reason for your answer.
\item Find an equation of the tangent to the curve at $P$.
\end{enumerate}\item The point $Q ( - 2,15 )$ also lies on the curve. Verify that $Q$ is a maximum point of the curve.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2014 Q3 [12]}}