CAIE P1 2020 November — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2020
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (extended problem with normals, stationary points, or further geometry)
DifficultyModerate -0.3 Part (a) requires applying the chain rule for related rates (dy/dt = dy/dx × dx/dt), a standard technique. Part (b) involves integrating fractional powers using the reverse power rule and finding the constant using the given point. Both parts are routine applications of standard techniques with no problem-solving insight required, making this slightly easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums

7 The point \(( 4,7 )\) lies on the curve \(y = \mathrm { f } ( x )\) and it is given that \(\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { - \frac { 1 } { 2 } } - 4 x ^ { - \frac { 3 } { 2 } }\).
  1. A point moves along the curve in such a way that the \(x\)-coordinate is increasing at a constant rate of 0.12 units per second. Find the rate of increase of the \(y\)-coordinate when \(x = 4\).
  2. Find the equation of the curve.

Question 7(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(f'(4) = \frac{5}{2}\)\*M1 Substituting 4 into \(f'(x)\)
\(\left(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\right) \rightarrow \left(\frac{dy}{dt}\right) = \frac{5}{2} \times 0.12\)DM1 Multiplies *their* \(f'(4)\) by 0.12
\(\frac{dy}{dt} = 0.3\)A1 OE
Question 7(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{6x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{4x^{-\frac{1}{2}}}{-\frac{1}{2}}(+c)\)B1 B1 B1 for each unsimplified integral
Uses \((4, 7)\) leading to \(c = -21\)M1 Uses \((4, 7)\) to find a \(c\) value
\(y\) or \(f(x) = 12x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} - 21\) or \(12\sqrt{x} + \frac{8}{\sqrt{x}} - 21\)A1 Need to see \(y\) or \(f(x) =\) somewhere in *their* solution and 12 and 8
## Question 7(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(4) = \frac{5}{2}$ | \*M1 | Substituting 4 into $f'(x)$ |
| $\left(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\right) \rightarrow \left(\frac{dy}{dt}\right) = \frac{5}{2} \times 0.12$ | DM1 | Multiplies *their* $f'(4)$ by 0.12 |
| $\frac{dy}{dt} = 0.3$ | A1 | OE |

## Question 7(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{6x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{4x^{-\frac{1}{2}}}{-\frac{1}{2}}(+c)$ | B1 B1 | B1 for each unsimplified integral |
| Uses $(4, 7)$ leading to $c = -21$ | M1 | Uses $(4, 7)$ to find a $c$ value |
| $y$ or $f(x) = 12x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} - 21$ or $12\sqrt{x} + \frac{8}{\sqrt{x}} - 21$ | A1 | Need to see $y$ or $f(x) =$ somewhere in *their* solution and 12 and 8 |

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7 The point $( 4,7 )$ lies on the curve $y = \mathrm { f } ( x )$ and it is given that $\mathrm { f } ^ { \prime } ( x ) = 6 x ^ { - \frac { 1 } { 2 } } - 4 x ^ { - \frac { 3 } { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item A point moves along the curve in such a way that the $x$-coordinate is increasing at a constant rate of 0.12 units per second.

Find the rate of increase of the $y$-coordinate when $x = 4$.
\item Find the equation of the curve.
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2020 Q7 [7]}}