| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Show/verify a given line is a tangent |
| Difficulty | Standard +0.8 Part (a) is routine differentiation and tangent-finding (standard AS calculus). Part (b) requires finding where this tangent line is perpendicular to the curve elsewhere—this involves setting up and solving a more complex equation relating gradient conditions at two different points, which goes beyond standard exercises and requires problem-solving insight, making it moderately challenging for AS level. |
| Spec | 1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| 10(a) | Rewrites the equation as |
| Answer | Marks | Guidance |
|---|---|---|
| √2 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| is negative. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| correct gradient. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| tangent to curve, any form. | 1.1b | A1 |
| Subtotal | 4 | 4 4 |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 10(b) | Eliminates y for their tangent |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑑𝑑𝑥𝑥 | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 1.1a | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| of x = –1 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| intersection point. | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| equa√ti2on found in part (a). OE | 2.1 | R1 |
| Subtotal | 5 | |
| Question 10 Total | 9 | |
| Q | Marking instructions | AO |
Question 10:
--- 10(a) ---
10(a) | Rewrites the equation as
y = x –2
PI by correct derivative.
√2 | 1.1b | B1 | y = x –2
√2
= –
𝑑𝑑𝑑𝑑 2√2
3
𝑑𝑑𝑥𝑥
𝑥𝑥
Grad at (2, ) = – = –
√2 2√2 √2
4
8 4
Tangent at (2, ) is
√2
y – = – (x4 – 2)
√2 √2
y 4= –4
3√2 𝑥𝑥√2
Differentiates with their power of
x correct provided original power
is negative. | 1.1a | M1
Substitutes x = 2 to obtain
correct gradient. | 1.1b | A1
Obtains correct equation of
tangent to curve, any form. | 1.1b | A1
Subtotal | 4 | 4 4
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b) ---
10(b) | Eliminates y for their tangent
and the given curve to find other
intersection point.
Or
Equates to the gradient of the
𝑑𝑑𝑑𝑑
perpendicular to their tangent.
𝑑𝑑𝑥𝑥 | 3.1a | M1 | Meets y = when
√2
2
𝑥𝑥
= –
√2 3√2 𝑥𝑥√2
2
𝑥𝑥 4 4
= –
1 3 𝑥𝑥
2
𝑥𝑥 4 4
x3 – 3x2 + 4 = 0
(x – 2)2(x + 1) = 0
Other intersection is at x = –1
= =
𝑑𝑑𝑑𝑑 −2√2
3
(−1) 2√2
𝑑𝑑𝑥𝑥
(– ) = –1
√2
Perpend 2 i √ cu 2 la × r to c4urve so normal
Simplifies to obtain correct cubic
equation.
PI by x = –1
Or
Obtains correct equation
– = (OE)
2√2
3 | 1.1a | A1
𝑥𝑥 2√2
Finds other intersection value
of x = –1 | 1.1b | A1
Substitutes x = -1 to obtain the
correct gradient at the other
intersection point.
Or
Obtains y = at the other
intersection point. | 1.1b | B1
√2
Completes a reasoned
argument to show the required
result using the perpendicular
gradients condition.
Or
Completes argument by finding
the equation of the line with
gradient – passing through
√2
(-1 , ) and verifying that this
4
equation is identical to the
equa√ti2on found in part (a). OE | 2.1 | R1
Subtotal | 5
Question 10 Total | 9
Q | Marking instructions | AO | Marks | Typical solutionCurve $C$ has equation $y = \frac{\sqrt{2}}{x^2}$
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the tangent to $C$ at the point $\left(2, \frac{\sqrt{2}}{4}\right)$
[4 marks]
\item Show that the tangent to $C$ at the point $\left(2, \frac{\sqrt{2}}{4}\right)$ is also a normal to the curve at a different point.
\includegraphics{figure_10}
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2022 Q10 [9]}}