Standard +0.8 This is a multi-step problem requiring: finding the derivative, evaluating at x=2, determining both tangent and normal equations, finding their x-intercepts, and calculating a triangular area. While each individual step is standard A-level technique, the combination of multiple concepts (differentiation, perpendicular gradients, coordinate geometry, area calculation) and the need to organize the solution coherently makes this moderately challenging—harder than routine single-concept questions but not requiring novel mathematical insight.
10 In this question you must show detailed reasoning.
The equation of a curve is \(y = \frac { x ^ { 2 } } { 4 } + \frac { 2 } { x } + 1\). A tangent and a normal to the curve are drawn at the point where \(x = 2\).
Calculate the area bounded by the tangent, the normal and the \(x\)-axis.
\section*{END OF QUESTION PAPER}
equation of tangent; giving \(y = \frac{1}{2}x + 2\)
\(y - 3 = -2(x - 2)\)
M1 [1.1]
and equation of normal; FT their tangent gradient; giving \(y = -2x + 7\)
intercepts are \((-4, 0)\)
A1 [1.1]
and \((3.5, 0)\)
A1 [1.1]
Area \(= \frac{1}{2} \times 7\frac{1}{2} \times 3\)
M1 [2.1]
OR \(\frac{1}{2} \times \sqrt{(6^2+3^2)} \times \sqrt{(3^2+1.5^2)}\); lengths of tangent and normal. May see \(3\sqrt{5}\) and \(\frac{3}{2}\sqrt{5}\) — or integration (complete method)
\(\frac{45}{4}\) cao
A1 [1.1]
or 11.25
[10 marks total]
## Question 10:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2, y = 3$ | **B1** [1.1] | soi |
| $\frac{dy}{dx} = \frac{2x}{4} - \frac{2}{x^2}$ oe | **M1** [2.1] | Differentiation with 1 term correct |
| | **A1** [1.1] | |
| Evaluation of their derivative at $x = 2$ | **M1** [1.1] | NB $\frac{1}{2}$ |
| $y - 3 = \frac{1}{2}(x - 2)$ | **M1** [2.4] | equation of tangent; giving $y = \frac{1}{2}x + 2$ |
| $y - 3 = -2(x - 2)$ | **M1** [1.1] | and equation of normal; FT their tangent gradient; giving $y = -2x + 7$ |
| intercepts are $(-4, 0)$ | **A1** [1.1] | |
| and $(3.5, 0)$ | **A1** [1.1] | |
| Area $= \frac{1}{2} \times 7\frac{1}{2} \times 3$ | **M1** [2.1] | OR $\frac{1}{2} \times \sqrt{(6^2+3^2)} \times \sqrt{(3^2+1.5^2)}$; lengths of tangent and normal. May see $3\sqrt{5}$ and $\frac{3}{2}\sqrt{5}$ — or integration (complete method) |
| $\frac{45}{4}$ cao | **A1** [1.1] | or 11.25 |
**[10 marks total]**
10 In this question you must show detailed reasoning.
The equation of a curve is $y = \frac { x ^ { 2 } } { 4 } + \frac { 2 } { x } + 1$. A tangent and a normal to the curve are drawn at the point where $x = 2$.
Calculate the area bounded by the tangent, the normal and the $x$-axis.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2019 Q10 [10]}}