| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Normal or tangent line problems |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring basic power rule integration with a linear substitution, plus finding a tangent line using point-slope form. Both parts are standard textbook exercises with no conceptual challenges—part (a) requires only substituting x=6 into the derivative and using y-y₁=m(x-x₁), while part (b) involves routine integration of (ax+b)^n and finding the constant using the given point. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(12\!\left(\dfrac{1}{2}\times6-1\right)^{-4}\left[=12(2)^{-4}=\dfrac{3}{4}\right]\) | M1 | Substitute \(x=6\) into \(\dfrac{dy}{dx}\); SOI by gradient \(\dfrac{3}{4}\) used |
| \(y-4=\dfrac{3}{4}(x-6)\), OR evaluates \(c=-\dfrac{1}{2}\) | A1 | OE e.g. \(y=\dfrac{3}{4}x-\dfrac{1}{2}\) or evaluates \(c\) in \(y=\dfrac{3}{4}x+c\) using \((6,4)\) and gradient \(\dfrac{3}{4}\); ISW |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([y=]\frac{12(\frac{1}{2}x-1)^{-3}}{-3} \div \frac{1}{2} [= -8(\frac{1}{2}x-1)^{-3}]\) | B2, 1, 0 | |
| \(4 = \frac{12\times(\frac{1}{2}\times6-1)^{-3}}{\frac{1}{2}\times-3} + c \Rightarrow 4 = -8\times2^{-3}+c \Rightarrow c=[5]\) | M1 | Must have \(+c\). Substitute \(y=4\), \(x=6\) and solve for \(c\) in an integrated expression. May be unsimplified. |
| \([y=]-8(\frac{1}{2}x-1)^{-3}+5\) | A1 | OE Must see '\(y=\)' or '\(f(x)=\)' in the working. |
| 4 |
**Question 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $12\!\left(\dfrac{1}{2}\times6-1\right)^{-4}\left[=12(2)^{-4}=\dfrac{3}{4}\right]$ | M1 | Substitute $x=6$ into $\dfrac{dy}{dx}$; SOI by gradient $\dfrac{3}{4}$ used |
| $y-4=\dfrac{3}{4}(x-6)$, OR evaluates $c=-\dfrac{1}{2}$ | A1 | OE e.g. $y=\dfrac{3}{4}x-\dfrac{1}{2}$ or evaluates $c$ in $y=\dfrac{3}{4}x+c$ using $(6,4)$ and gradient $\dfrac{3}{4}$; ISW |
| **Total: 2** | | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[y=]\frac{12(\frac{1}{2}x-1)^{-3}}{-3} \div \frac{1}{2} [= -8(\frac{1}{2}x-1)^{-3}]$ | **B2, 1, 0** | |
| $4 = \frac{12\times(\frac{1}{2}\times6-1)^{-3}}{\frac{1}{2}\times-3} + c \Rightarrow 4 = -8\times2^{-3}+c \Rightarrow c=[5]$ | **M1** | Must have $+c$. Substitute $y=4$, $x=6$ and solve for $c$ in an integrated expression. May be unsimplified. |
| $[y=]-8(\frac{1}{2}x-1)^{-3}+5$ | **A1** | OE Must see '$y=$' or '$f(x)=$' in the working. |
| | **4** | |
---2 The equation of a curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 12 \left( \frac { 1 } { 2 } x - 1 \right) ^ { - 4 }$. It is given that the curve passes through the point $P ( 6,4 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve at $P$.
\item Find the equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q2 [6]}}