CAIE P1 2020 November — Question 6 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2020
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeTangent with specified gradient
DifficultyStandard +0.3 This is a straightforward chain rule differentiation followed by solving a simple equation. Students differentiate to get dy/dx = -x/√(25-x²), set equal to 4/3, and solve for x. The algebra is routine and the question requires only standard technique application with no conceptual challenges.
Spec1.07i Differentiate x^n: for rational n and sums1.07q Product and quotient rules: differentiation
6 The equation of a curve is \(y = 2 + \sqrt { 25 - x ^ { 2 } }\).
Find the coordinates of the point on the curve at which the gradient is \(\frac { 4 } { 3 }\).
Question 6:
AnswerMarks Guidance
\(\frac{dy}{dx} = \left[\frac{1}{2}(25-x^2)^{-1/2}\right] \times [-2x]\)B1 B1
\(\frac{-x}{(25-x^2)^{1/2}} = \frac{4}{3} \rightarrow \frac{x^2}{25-x^2} = \frac{16}{9}\)M1 Set \(= \frac{4}{3}\) and square both sides
\(16(25-x^2) = 9x^2 \rightarrow 25x^2 = 400 \rightarrow x = (\pm)4\)A1
When \(x = -4\), \(y = 5 \rightarrow (-4, 5)\)A1 
## Question 6:

| $\frac{dy}{dx} = \left[\frac{1}{2}(25-x^2)^{-1/2}\right] \times [-2x]$ | B1 B1 | |
|---|---|---|
| $\frac{-x}{(25-x^2)^{1/2}} = \frac{4}{3} \rightarrow \frac{x^2}{25-x^2} = \frac{16}{9}$ | M1 | Set $= \frac{4}{3}$ and square both sides |
| $16(25-x^2) = 9x^2 \rightarrow 25x^2 = 400 \rightarrow x = (\pm)4$ | A1 | |
| When $x = -4$, $y = 5 \rightarrow (-4, 5)$ | A1 | |

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6 The equation of a curve is $y = 2 + \sqrt { 25 - x ^ { 2 } }$.\\
Find the coordinates of the point on the curve at which the gradient is $\frac { 4 } { 3 }$.\\

\hfill \mbox{\textit{CAIE P1 2020 Q6 [5]}}
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