OCR C3 2006 June — Question 1 5 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2006
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeEquation of tangent line
DifficultyModerate -0.8 This is a straightforward application of the chain rule to find dy/dx, evaluate at the given point, then use y - y₁ = m(x - x₁). The chain rule here is simple (power of a linear function), and all steps are routine with no problem-solving required. Easier than average for A-level.
Spec1.07m Tangents and normals: gradient and equations

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

AnswerMarks Guidance
Differentiate to obtain \(k(4x+1)^{-\frac{1}{2}}\)M1 any non-zero constant \(k\)
Obtain \(2(4x+1)^{-\frac{1}{2}}\)A1 or equiv, perhaps unsimplified
Obtain \(\frac{2}{3}\) for value of first derivativeA1 or unsimplified equiv
Attempt equation of tangent through \((2, 3)\)M1 using numerical value of first derivative provided derivative is of form \(k'(4x+1)^n\)
Obtain \(y = \frac{2}{3}x + \frac{5}{3}\) or \(2x - 3y + 5 = 0\)A1 5 or equiv involving 3 terms
Differentiate to obtain $k(4x+1)^{-\frac{1}{2}}$ | M1 | any non-zero constant $k$

Obtain $2(4x+1)^{-\frac{1}{2}}$ | A1 | or equiv, perhaps unsimplified

Obtain $\frac{2}{3}$ for value of first derivative | A1 | or unsimplified equiv

Attempt equation of tangent through $(2, 3)$ | M1 | using numerical value of first derivative provided derivative is of form $k'(4x+1)^n$

Obtain $y = \frac{2}{3}x + \frac{5}{3}$ or $2x - 3y + 5 = 0$ | A1 5 | or equiv involving 3 terms
1 Find the equation of the tangent to the curve $y = \sqrt { 4 x + 1 }$ at the point ( 2,3 ).

\hfill \mbox{\textit{OCR C3 2006 Q1 [5]}}