OCR MEI C2 2011 June — Question 5 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyModerate -0.8 This is a straightforward application of differentiation requiring finding dy/dx, evaluating at a point, finding the negative reciprocal for the normal gradient, and using y - y₁ = m(x - x₁). All steps are routine with simple arithmetic (x = 1/2 makes calculations easy). This is easier than average as it's a single-technique question with no problem-solving element.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations
5 Find the equation of the normal to the curve \(y = 8 x ^ { 4 } + 4\) at the point where \(x = \frac { 1 } { 2 }\).
AnswerMarks Guidance
\(\int \frac{dy}{dx} = \int 3kx^3\) c.a.o.M1
substitution of \(x = \frac{1}{2}\) in their \(\frac{dy}{dx}\)M1 \([= 4]\)
\(\text{grad normal} = \frac{-1}{\text{their 4}}\)M1 their 4 must be obtained by calculus
when \(x = \frac{1}{2}, y = 4\frac{1}{2}\) o.e.B1
\(y - 4\frac{1}{2} = -\frac{1}{4}(x - \frac{1}{2})\) i.s.wA1 \(y = -\frac{1}{4}x + 4\frac{3}{8}\) o.e. 
$\int \frac{dy}{dx} = \int 3kx^3$ c.a.o. | M1 |
substitution of $x = \frac{1}{2}$ in their $\frac{dy}{dx}$ | M1 | $[= 4]$ | must see $kx^3$
$\text{grad normal} = \frac{-1}{\text{their 4}}$ | M1 | their 4 must be obtained by calculus
when $x = \frac{1}{2}, y = 4\frac{1}{2}$ o.e. | B1 |
$y - 4\frac{1}{2} = -\frac{1}{4}(x - \frac{1}{2})$ i.s.w | A1 | $y = -\frac{1}{4}x + 4\frac{3}{8}$ o.e.
5 Find the equation of the normal to the curve $y = 8 x ^ { 4 } + 4$ at the point where $x = \frac { 1 } { 2 }$.

\hfill \mbox{\textit{OCR MEI C2 2011 Q5 [5]}}
This paper (13 questions)
View full paper
Q1 3 Q2 3 Q3 5 Q4 3 Q5 5 Q6 5 Q7 4 Q8 3 Q9 3 Q10 2 Q11 11 Q12 17 Q13 12