OCR C3 2015 June — Question 1 5 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeFind equation of tangent
DifficultyModerate -0.3 This is a straightforward application of the quotient rule followed by finding a tangent line equation. While it requires multiple steps (differentiate using quotient rule, substitute x=2 to find gradient, use point-slope form), these are all standard procedures with no conceptual challenges. Slightly easier than average due to being a direct application problem with clean arithmetic.
Spec1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

1 Find the equation of the tangent to the curve \(y = \frac { 5 x + 4 } { 3 x - 8 }\) at the point \(( 2 , - 7 )\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Attempt use of quotient rule or, after adjustment, product rule\*M1 For M1 allow one slip in numerator but must be minus sign in numerator and square of \(3x-8\) in denominator; allow M1 for numerator the wrong way round. For product rule attempt, \*M1 for \(k_1(3x-8)^{-1}+k_2(5x+4)(3x-8)^{-2}\) form and A1 for correct constants 5 and \(-3\)
Obtain \(\frac{5(3x-8)-3(5x+4)}{(3x-8)^2}\) or equivA1 Allow if missing brackets implied by subsequent simplification or calculation
Substitute 2 to obtain \(-13\) or equivA1
Attempt to find equation of tangentM1 Dep \*M; equation of tangent not normal
Obtain \(y=-13x+19\) or \(13x+y-19=0\)A1 Or similarly simplified equiv with 3 non-zero terms
[5]
# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of quotient rule or, after adjustment, product rule | \*M1 | For M1 allow one slip in numerator but must be minus sign in numerator and square of $3x-8$ in denominator; allow M1 for numerator the wrong way round. For product rule attempt, \*M1 for $k_1(3x-8)^{-1}+k_2(5x+4)(3x-8)^{-2}$ form and A1 for correct constants 5 and $-3$ |
| Obtain $\frac{5(3x-8)-3(5x+4)}{(3x-8)^2}$ or equiv | A1 | Allow if missing brackets implied by subsequent simplification or calculation |
| Substitute 2 to obtain $-13$ or equiv | A1 | |
| Attempt to find equation of tangent | M1 | Dep \*M; equation of tangent not normal |
| Obtain $y=-13x+19$ or $13x+y-19=0$ | A1 | Or similarly simplified equiv with 3 non-zero terms |
| **[5]** | | |

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1 Find the equation of the tangent to the curve $y = \frac { 5 x + 4 } { 3 x - 8 }$ at the point $( 2 , - 7 )$.

\hfill \mbox{\textit{OCR C3 2015 Q1 [5]}}