| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find normal line equation |
| Difficulty | Moderate -0.3 Part (i) requires finding a derivative of a square root function, evaluating it, finding the perpendicular gradient, and writing a line equation—all standard techniques. Part (ii) involves integrating the square root function and subtracting a triangle area, which is routine A-level integration. This is slightly easier than average due to straightforward algebraic manipulation and standard procedures throughout. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2} \times (5x-1)^{-\frac{1}{2}} \times 5 \quad \left(= \frac{5}{6}\right)\) | B1 B1 | B1 Without \(\times 5\) B1 \(\times 5\) of an attempt at differentiation |
| \(m\) of normal \(= -\frac{6}{5}\) | M1 | Uses \(m_1 m_2 = -1\) with their numeric value from their \(dy/dx\) |
| Equation of normal \(y - 3 = -\frac{6}{5}(x-2)\) OE or \(5y + 6x = 27\) or \(y = -\frac{6}{5}x + \frac{27}{5}\) | A1 | Unsimplified. Can use \(y = mx + c\) to get \(c = 5.4\) ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER: \(\int \sqrt{5x-1}\, dx = \frac{(5x-1)^{\frac{3}{2}}}{\frac{3}{2}} \div 5\) | (B1 | Correct expression without \(\div 5\) |
| B1 | For dividing an attempt at integration of \(y\) by 5 | |
| Limits from \(\frac{1}{5}\) to 2 used \(\rightarrow 3.6\) or \(\frac{18}{5}\) OE | M1 A1 | Using \(\frac{1}{5}\) and 2 to evaluate an integrand (may be \(\int y^2\)) |
| Normal crosses \(x\)-axis when \(y = 0 \rightarrow x = 4\frac{1}{2}\) | M1 | Uses their equation of normal, NOT tangent |
| Area of triangle \(= 3.75\) or \(\frac{15}{4}\) OE | A1 | This can be obtained by integration |
| Total area \(= 3.6 + 3.75 = 7.35\), \(\frac{147}{20}\) OE | A1 | |
| OR: \(\int \frac{1}{5}(y^2+1)\,dy = \frac{1}{5}\left(\frac{y^3}{3} + y\right)\) | (B2, 1, 0) | \(-1\) each error or omission |
| Limits from 0 to 3 used \(\rightarrow 2.4\) or \(\frac{12}{5}\) OE | M1 A1 | Using 0 and 3 to evaluate an integrand |
| Uses their equation of normal, NOT tangent | M1 | Either to find side length for trapezium or attempt at integrating between 0 and 3 |
| Area of trapezium \(= \frac{1}{2}(2 + 4\frac{1}{2}) \times 3 = \frac{39}{4}\) or \(9\frac{3}{4}\) | A1 | This can be obtained by integration |
| Shaded area \(= \frac{39}{4} - \frac{12}{5} = 7.35, \frac{147}{20}\) OE | A1 | |
| 7 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{2} \times (5x-1)^{-\frac{1}{2}} \times 5 \quad \left(= \frac{5}{6}\right)$ | B1 B1 | **B1** Without $\times 5$ **B1** $\times 5$ of an attempt at differentiation |
| $m$ of normal $= -\frac{6}{5}$ | M1 | Uses $m_1 m_2 = -1$ with their numeric value from their $dy/dx$ |
| Equation of normal $y - 3 = -\frac{6}{5}(x-2)$ OE or $5y + 6x = 27$ or $y = -\frac{6}{5}x + \frac{27}{5}$ | A1 | Unsimplified. Can use $y = mx + c$ to get $c = 5.4$ ISW |
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** $\int \sqrt{5x-1}\, dx = \frac{(5x-1)^{\frac{3}{2}}}{\frac{3}{2}} \div 5$ | (B1 | Correct expression without $\div 5$ |
| | B1 | For dividing an attempt at integration of $y$ by 5 |
| Limits from $\frac{1}{5}$ to 2 used $\rightarrow 3.6$ or $\frac{18}{5}$ OE | M1 A1 | Using $\frac{1}{5}$ and 2 to evaluate an integrand (may be $\int y^2$) |
| Normal crosses $x$-axis when $y = 0 \rightarrow x = 4\frac{1}{2}$ | M1 | Uses their equation of normal, NOT tangent |
| Area of triangle $= 3.75$ or $\frac{15}{4}$ OE | A1 | This can be obtained by integration |
| Total area $= 3.6 + 3.75 = 7.35$, $\frac{147}{20}$ OE | A1 | |
| **OR:** $\int \frac{1}{5}(y^2+1)\,dy = \frac{1}{5}\left(\frac{y^3}{3} + y\right)$ | (B2, 1, 0) | $-1$ each error or omission |
| Limits from 0 to 3 used $\rightarrow 2.4$ or $\frac{12}{5}$ OE | M1 A1 | Using 0 and 3 to evaluate an integrand |
| Uses their equation of normal, NOT tangent | M1 | Either to find side length for trapezium or attempt at integrating between 0 and 3 |
| Area of trapezium $= \frac{1}{2}(2 + 4\frac{1}{2}) \times 3 = \frac{39}{4}$ or $9\frac{3}{4}$ | A1 | This can be obtained by integration |
| Shaded area $= \frac{39}{4} - \frac{12}{5} = 7.35, \frac{147}{20}$ OE | A1 | |
| | **7** | |10\\
\includegraphics[max width=\textwidth, alt={}, center]{518bb805-5b14-4b41-94fd-38a31a90c218-18_551_689_260_726}
The diagram shows part of the curve $y = \sqrt { } ( 5 x - 1 )$ and the normal to the curve at the point $P ( 2,3 )$. This normal meets the $x$-axis at $Q$.\\
(i) Find the equation of the normal at $P$.\\
(ii) Find, showing all necessary working, the area of the shaded region.\\
\hfill \mbox{\textit{CAIE P1 2017 Q10 [11]}}