Edexcel C1 2012 June — Question 7 8 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2012
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind tangent given derivative expression
DifficultyModerate -0.3 This is a straightforward C1 question requiring basic integration and tangent line calculation. Part (a) involves substituting x=4 into the given derivative to find the gradient, then using point-slope form. Part (b) requires integrating a polynomial and x^(-1/2) term, then finding the constant using the given point. Both parts are standard textbook exercises with no problem-solving insight required, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums

7. The point \(P ( 4 , - 1 )\) lies on the curve \(C\) with equation \(y = \mathrm { f } ( x ) , x > 0\), and $$f ^ { \prime } ( x ) = \frac { 1 } { 2 } x - \frac { 6 } { \sqrt { } x } + 3$$
  1. Find the equation of the tangent to \(C\) at the point \(P\), giving your answer in the form \(y = m x + c\), where \(m\) and \(c\) are integers.
  2. Find \(\mathrm { f } ( x )\).

7. The point $P ( 4 , - 1 )$ lies on the curve $C$ with equation $y = \mathrm { f } ( x ) , x > 0$, and

$$f ^ { \prime } ( x ) = \frac { 1 } { 2 } x - \frac { 6 } { \sqrt { } x } + 3$$
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to $C$ at the point $P$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are integers.
\item Find $\mathrm { f } ( x )$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2012 Q7 [8]}}