| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard calculus techniques: differentiation using chain rule, finding normal equation, locating and classifying a stationary point, and computing a definite integral. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average but still requiring competent execution across multiple techniques. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1\) | B1 B1 | B1 all correct with 1 error, B2 if all correct |
| Gradient of tangent \(= -\frac{1}{4}\) and Gradient of normal \(= 4\) | *M1 | Substituting \(x=4\) into a differentiated expression and using \(m_1 m_2 = -1\) |
| Equation of line is \((y-4) = 4(x-4)\) or evaluate \(c\) | DM1 | With \((4, 4)\) and *their* gradient of normal |
| So \(y = 4x - 12\) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3(3x+4)^{-0.5} - 1 = 0\) | M1 | Setting *their* \(\frac{dy}{dx} = 0\) |
| Solving as far as \(x =\) | M1 | Where \(\frac{dy}{dx}\) contains \(a(bx+c)^{-0.5}\), \(a\), \(b\), \(c\) any values |
| \(x = \frac{5}{3}\), \(y = 2\left(3 \times \frac{5}{3}+4\right)^{0.5} - \frac{5}{3} = \frac{13}{3}\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = -\frac{9}{2}(3x+4)^{-1.5}\) | M1 | Differentiating *their* \(\frac{dy}{dx}\) OR checking \(\frac{dy}{dx}\) to find \(+\)ve and \(-\)ve either side of their \(x = \frac{5}{3}\) |
| At \(x = \frac{5}{3}\), \(\frac{d^2y}{dx^2}\) is negative so the point is a maximum | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Area} = \left[\int 2(3x+4)^{0.5} - x \, dx =\right] \frac{4}{9}(3x+4)^{1.5} - \frac{1}{2}x^2\) | B1 B1 | B1 for each correct term (unsimplified) |
| \(\left(\frac{4}{9}(16)^{1.5} - \frac{1}{2}(4)^2\right) - \frac{4}{9}(4)^{1.5} = \frac{256}{9} - 8 - \frac{32}{9}\) | M1 | Substituting limits 0 and 4 into an expression obtained by integrating \(y\) |
| \(16\dfrac{8}{9}\) | A1 | Or \(\dfrac{152}{9}\) |
| 4 |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1$ | B1 B1 | B1 all correct with 1 error, B2 if all correct |
| Gradient of tangent $= -\frac{1}{4}$ and Gradient of normal $= 4$ | *M1 | Substituting $x=4$ into a differentiated expression and using $m_1 m_2 = -1$ |
| Equation of line is $(y-4) = 4(x-4)$ **or** evaluate $c$ | DM1 | With $(4, 4)$ and *their* gradient of normal |
| So $y = 4x - 12$ | A1 | |
| | **5** | |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3(3x+4)^{-0.5} - 1 = 0$ | M1 | Setting *their* $\frac{dy}{dx} = 0$ |
| Solving as far as $x =$ | M1 | Where $\frac{dy}{dx}$ contains $a(bx+c)^{-0.5}$, $a$, $b$, $c$ any values |
| $x = \frac{5}{3}$, $y = 2\left(3 \times \frac{5}{3}+4\right)^{0.5} - \frac{5}{3} = \frac{13}{3}$ | A1 | |
| | **3** | |
---
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = -\frac{9}{2}(3x+4)^{-1.5}$ | M1 | Differentiating *their* $\frac{dy}{dx}$ OR checking $\frac{dy}{dx}$ to find $+$ve and $-$ve either side of their $x = \frac{5}{3}$ |
| At $x = \frac{5}{3}$, $\frac{d^2y}{dx^2}$ is negative so the point is a maximum | A1 | |
| | **2** | |
## Question 11(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Area} = \left[\int 2(3x+4)^{0.5} - x \, dx =\right] \frac{4}{9}(3x+4)^{1.5} - \frac{1}{2}x^2$ | **B1 B1** | B1 for each correct term (unsimplified) |
| $\left(\frac{4}{9}(16)^{1.5} - \frac{1}{2}(4)^2\right) - \frac{4}{9}(4)^{1.5} = \frac{256}{9} - 8 - \frac{32}{9}$ | **M1** | Substituting limits 0 and 4 into an expression obtained by integrating $y$ |
| $16\dfrac{8}{9}$ | **A1** | Or $\dfrac{152}{9}$ |
| | **4** | |11 The equation of a curve is $y = 2 \sqrt { 3 x + 4 } - x$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the normal to the curve at the point (4,4), giving your answer in the form $y = m x + c$.
\item Find the coordinates of the stationary point.
\item Determine the nature of the stationary point.
\item Find the exact area of the region bounded by the curve, the $x$-axis and the lines $x = 0$ and $x = 4$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q11 [14]}}