OCR H240/01 2018 September — Question 6 7 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2018
SessionSeptember
Marks7
TopicProduct & Quotient Rules
TypeFind equation of tangent
DifficultyStandard +0.3 This is a straightforward tangent line question requiring quotient rule and chain rule differentiation, then point-slope form. While it involves multiple differentiation techniques and algebraic manipulation to reach the given answer, it follows a completely standard procedure with no problem-solving insight required. The 'show that' format makes it slightly easier than finding the equation independently. Slightly above average difficulty due to the algebraic complexity of combining the quotient rule term with the chain rule term and simplifying to the exact form given.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation
6 In this question you must show detailed reasoning. A curve has equation \(y = \frac { 2 x } { 3 x - 1 } + \sqrt { 5 x + 1 }\). Show that the equation of the tangent to the curve at the point where \(x = 3\) is \(19 x - 32 y + 95 = 0\).
AnswerMarks Guidance
DR\(\frac{2(3x-1) - 6x}{(3x-1)^2}\) M1
\(\frac{5}{2}(5x+1)^{-\frac{1}{2}}\)M1 Attempt to differentiate second term (Obtain \(k(5x+1)^{-\frac{1}{2}}\))
\(\frac{dy}{dx} = \frac{-2}{(3x-1)^2} + \frac{5}{2}(5x+1)^{-1}\)A1 Fully correct derivative (Allow unsimplified)
At \(x = 3\), \(\frac{dy}{dx} = -\frac{2}{64} + \frac{5}{8} = \frac{19}{32}\)M1 Attempt gradient at \(x = 3\)
At \(x = 3\), \(y = \frac{10}{4}\)B1 Correct y-coordinate
\(y - \frac{10}{4} = \frac{19}{32}(x - 3)\)M1 Attempt equation of line with their y-coordinate and gradient
\(32y - 152 = 19x - 57\)A1 Rearrange to given form (At least one line of working seen)
\(19x - 32y + 95 = 0\)AG

Total: [7]

DR | $\frac{2(3x-1) - 6x}{(3x-1)^2}$ | M1 | Attempt to differentiate first term using quotient rule or equiv
| $\frac{5}{2}(5x+1)^{-\frac{1}{2}}$ | M1 | Attempt to differentiate second term (Obtain $k(5x+1)^{-\frac{1}{2}}$)
| $\frac{dy}{dx} = \frac{-2}{(3x-1)^2} + \frac{5}{2}(5x+1)^{-1}$ | A1 | Fully correct derivative (Allow unsimplified)
| At $x = 3$, $\frac{dy}{dx} = -\frac{2}{64} + \frac{5}{8} = \frac{19}{32}$ | M1 | Attempt gradient at $x = 3$
| At $x = 3$, $y = \frac{10}{4}$ | B1 | Correct y-coordinate
| $y - \frac{10}{4} = \frac{19}{32}(x - 3)$ | M1 | Attempt equation of line with their y-coordinate and gradient
| $32y - 152 = 19x - 57$ | A1 | Rearrange to given form (At least one line of working seen)
| $19x - 32y + 95 = 0$ | AG | |
Total: [7]
6 In this question you must show detailed reasoning.
A curve has equation $y = \frac { 2 x } { 3 x - 1 } + \sqrt { 5 x + 1 }$. Show that the equation of the tangent to the curve at the point where $x = 3$ is $19 x - 32 y + 95 = 0$.

\hfill \mbox{\textit{OCR H240/01 2018 Q6 [7]}}
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