| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | March |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques: solving a cubic equation to find intersection points, using differentiation to find a tangent parallel to a given line, and finding a normal equation. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| 10(i) | 4x1/2 = x+3 → | |
| (x1/2)2 −4x1/2 +3 (=0 ) OR 16x= x2 +6x+9 | M1 | Eliminate y from the 2 equations and then: |
| Answer | Marks | Guidance |
|---|---|---|
| x1/2 =1 or 3 x2 −10x+9 (=0 ) | A1 | If in 1st method x1/2 becomes x, allow only M1 unless |
| Answer | Marks | Guidance |
|---|---|---|
| x = 1 or 9 | A1 | |
| y=4 or1 2 | A1ft | Ft from their x values |
| Answer | Marks |
|---|---|
| AB2 9−1 12−4 | M1 |
| AB= 128 or 8 2 oe or 11.3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10(ii) | dy/dx = 2x−1/2 | B1 |
| 2x−1/2 = 1 | M1 | Set their derivative = their gradient of AB and attempt |
| Answer | Marks | Guidance |
|---|---|---|
| (4, 8) | A1 | Alternative method without calculus: |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 10(iii) | Equation of normal is y−8=−1 ( x−4 ) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Eliminate y (or x) → −x+12=x+3 or y−3=12− y | M1 | May use their equation of AB |
| (4½, 7½) | A1 |
Question 10:
--- 10(i) ---
10(i) | 4x1/2 = x+3 →
(x1/2)2 −4x1/2 +3 (=0 ) OR 16x= x2 +6x+9 | M1 | Eliminate y from the 2 equations and then:
Either treat as quad in x1/2 OR square both sides and
RHS is 3-term
x1/2 =1 or 3 x2 −10x+9 (=0 ) | A1 | If in 1st method x1/2 becomes x, allow only M1 unless
subsequently squared
x = 1 or 9 | A1
y=4 or1 2 | A1ft | Ft from their x values
If the 2 solutions are found by trial substitution B1 for
the first coordinate and B3 for the second coordinate
=( )2 +( )2
AB2 9−1 12−4 | M1
AB= 128 or 8 2 oe or 11.3 | A1
6
--- 10(ii) ---
10(ii) | dy/dx = 2x−1/2 | B1
2x−1/2 = 1 | M1 | Set their derivative = their gradient of AB and attempt
to solve
(4, 8) | A1 | Alternative method without calculus:
M = 1, tangent is y = mx + c where m = 1 and meets
AB
y = 4x1/2 when 4x1/2 = x + c. This is a quadratic with
b2 = 4ac, so 16 – 4 ×1 ×(cid:1855)= 0 so c = 4 B1 Solving
4x1/2 = x + 4 gives x = 4 and y = 8 M1A1
3
Question | Answer | Marks | Guidance
--- 10(iii) ---
10(iii) | Equation of normal is y−8=−1 ( x−4 ) | M1 | Equation through their T and with gradient ‒1/their
gradient of AB. Expect y=−x+12,
Eliminate y (or x) → −x+12=x+3 or y−3=12− y | M1 | May use their equation of AB
(4½, 7½) | A1
3\includegraphics{figure_10}
The diagram shows the curve with equation $y = 4x^{\frac{1}{3}}$.
\begin{enumerate}[label=(\roman*)]
\item The straight line with equation $y = x + 3$ intersects the curve at points $A$ and $B$. Find the length of $AB$. [6]
\item The tangent to the curve at a point $T$ is parallel to $AB$. Find the coordinates of $T$. [3]
\item Find the coordinates of the point of intersection of the normal to the curve at $T$ with the line $AB$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q10 [12]}}