AQA C2 2015 June — Question 8 5 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent parallel to given line
DifficultyModerate -0.3 This is a straightforward C2 differentiation application requiring students to: (1) differentiate a power function, (2) rearrange a line to find its gradient, (3) equate gradients to find the x-coordinate, (4) find the y-coordinate, and (5) write the tangent equation. All steps are routine procedures with no conceptual challenges, making it slightly easier than average but still requiring multiple standard techniques.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations
8 The point \(A\) lies on the curve with equation \(y = x ^ { \frac { 1 } { 2 } }\). The tangent to this curve at \(A\) is parallel to the line \(3 y - 2 x = 1\). Find an equation of this tangent at \(A\).
[0pt] [5 marks]
Question 8:
The point \(A\) lies on \(y = x^{\frac{1}{2}}\), tangent parallel to \(3y - 2x = 1\). Find equation of tangent at \(A\).
AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}\)M1 A1 Differentiating
Gradient of line \(= \frac{2}{3}\), so \(\frac{1}{2}x^{-\frac{1}{2}} = \frac{2}{3}\)M1 Setting derivative equal to gradient of given line
\(x^{\frac{1}{2}} = \frac{3}{4}\), so \(x = \frac{9}{16}\)A1 Correct \(x\) value
\(y = \frac{3}{4}\), tangent: \(y - \frac{3}{4} = \frac{2}{3}(x - \frac{9}{16})\)
\(y = \frac{2}{3}x + \frac{3}{8}\) or equivalentA1 Correct equation 
## Question 8:

The point $A$ lies on $y = x^{\frac{1}{2}}$, tangent parallel to $3y - 2x = 1$. Find equation of tangent at $A$.

| $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ | M1 A1 | Differentiating |
|---|---|---|
| Gradient of line $= \frac{2}{3}$, so $\frac{1}{2}x^{-\frac{1}{2}} = \frac{2}{3}$ | M1 | Setting derivative equal to gradient of given line |
| $x^{\frac{1}{2}} = \frac{3}{4}$, so $x = \frac{9}{16}$ | A1 | Correct $x$ value |
| $y = \frac{3}{4}$, tangent: $y - \frac{3}{4} = \frac{2}{3}(x - \frac{9}{16})$ | | |
| $y = \frac{2}{3}x + \frac{3}{8}$ or equivalent | A1 | Correct equation |

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8 The point $A$ lies on the curve with equation $y = x ^ { \frac { 1 } { 2 } }$. The tangent to this curve at $A$ is parallel to the line $3 y - 2 x = 1$. Find an equation of this tangent at $A$.\\[0pt]
[5 marks]

\hfill \mbox{\textit{AQA C2 2015 Q8 [5]}}
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