OCR C1 — Question 9 10 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeNormal meets curve/axis — further geometry
DifficultyStandard +0.3 This is a standard C1 differentiation question requiring the quotient/power rule, finding a normal equation, and solving a simultaneous equation. While part (iii) involves solving a cubic, it factors easily. The techniques are routine and well-practiced, making it slightly easier than average for a multi-part calculus question.
Spec1.02q Use intersection points: of graphs to solve equations1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

A curve has the equation \(y = x + \frac{3}{x}\), \(x \neq 0\). The point \(P\) on the curve has \(x\)-coordinate \(1\).
  1. Show that the gradient of the curve at \(P\) is \(-2\). [3]
  2. Find an equation for the normal to the curve at \(P\), giving your answer in the form \(y = mx + c\). [3]
  3. Find the coordinates of the point where the normal to the curve at \(P\) intersects the curve again. [4]

(i)
AnswerMarks
\(\frac{dy}{dx} = 1 - 3x^{-2}\)M1 A1
grad \(= 1 - 3(1)^{-2} = 1 - 3 = -2\)A1
(ii)
AnswerMarks
\(x = 1\) ∴ \(y = 4\)M1
grad \(= \frac{-1}{-2} = \frac{1}{2}\)M1
∴ \(y - 4 = \frac{1}{2}(x - 1)\)M1
\(y = \frac{1}{2}x + \frac{7}{2}\)A1
(iii)
AnswerMarks
\(x + 3 = \frac{1}{2}x + \frac{7}{2}\)M1
\(2x^2 + 6 = x^2 + 7x\)M1
\(x^2 - 7x + 6 = 0\)M1
\((x - 1)(x - 6) = 0\)M1
\(x = 1\) (at \(P\)), \(6\)A1
∴ \((6, 6\frac{1}{2})\)A1
(10)
## (i)
$\frac{dy}{dx} = 1 - 3x^{-2}$ | M1 A1
grad $= 1 - 3(1)^{-2} = 1 - 3 = -2$ | A1

## (ii)
$x = 1$ ∴ $y = 4$ | M1
grad $= \frac{-1}{-2} = \frac{1}{2}$ | M1
∴ $y - 4 = \frac{1}{2}(x - 1)$ | M1
$y = \frac{1}{2}x + \frac{7}{2}$ | A1

## (iii)
$x + 3 = \frac{1}{2}x + \frac{7}{2}$ | M1
$2x^2 + 6 = x^2 + 7x$ | M1
$x^2 - 7x + 6 = 0$ | M1
$(x - 1)(x - 6) = 0$ | M1
$x = 1$ (at $P$), $6$ | A1
∴ $(6, 6\frac{1}{2})$ | A1
| (10)

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A curve has the equation $y = x + \frac{3}{x}$, $x \neq 0$.

The point $P$ on the curve has $x$-coordinate $1$.
\begin{enumerate}[label=(\roman*)]
\item Show that the gradient of the curve at $P$ is $-2$. [3]
\item Find an equation for the normal to the curve at $P$, giving your answer in the form $y = mx + c$. [3]
\item Find the coordinates of the point where the normal to the curve at $P$ intersects the curve again. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR C1  Q9 [10]}}