OCR C1 2008 June — Question 5 5 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2008
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind derivative after algebraic simplification (fractional/mixed powers)
DifficultyEasy -1.2 This is a straightforward differentiation question requiring rewriting √x as x^(1/2), applying the power rule to get dy/dx = 4x^(-1/2) + 1, then substituting x=9. It's simpler than average as it involves only basic power rule application with no problem-solving or multi-step reasoning required.
Spec1.07i Differentiate x^n: for rational n and sums
5 Find the gradient of the curve \(y = 8 \sqrt { x } + x\) at the point whose \(x\)-coordinate is 9 .
Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Attempt to differentiateM1 Attempt to differentiate
\(kx^{-\frac{1}{2}}\)A1
\(\frac{dy}{dx} = 4x^{-\frac{1}{2}} + 1\)A1
\(= 4\left(\frac{1}{\sqrt{9}}\right) + 1\)M1 Correct substitution of \(x = 9\) into their derivative
\(\frac{dy}{dx} = \frac{7}{3}\)A1 \(\frac{7}{3}\) only 
## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to differentiate | M1 | Attempt to differentiate |
| $kx^{-\frac{1}{2}}$ | A1 | |
| $\frac{dy}{dx} = 4x^{-\frac{1}{2}} + 1$ | A1 | |
| $= 4\left(\frac{1}{\sqrt{9}}\right) + 1$ | M1 | Correct substitution of $x = 9$ into their derivative |
| $\frac{dy}{dx} = \frac{7}{3}$ | A1 | $\frac{7}{3}$ only |

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5 Find the gradient of the curve $y = 8 \sqrt { x } + x$ at the point whose $x$-coordinate is 9 .

\hfill \mbox{\textit{OCR C1 2008 Q5 [5]}}
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Q1 4 Q2 3 Q3 4 Q4 5 Q5 5 Q6 6 Q7 7 Q8 10 Q9 14 Q10 14