| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Standard +0.3 This is a standard C2 differentiation question with routine steps: differentiate a simple power function, find a stationary point, determine a normal line equation, and apply a translation. All techniques are textbook exercises requiring no novel insight, though part (d) adds a minor geometric element that slightly elevates it above the most basic questions. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\frac{dy}{dx} = \frac{6}{2} x^{-0.5} - 1 = 3x^{-0.5} - 1\) | B2,1 |
| (b) | \(3x^{-0.5} - 1 = 0\) | M1 |
| (b) | \(3x^{-0.5} = 1, \quad x = 9\) | A1F |
| (b) | (y-coordinate of M is) 6 | A1 |
| (c) | At P(4,5) \(\frac{dy}{dx} = 3(4)^{-0.5} - 1 (=0.5)\) | M1 |
| (c) | Gradient of normal = \(-2\) | m1 |
| (c) | Eqn of normal: \(y - 5 = -2(x - 4)\) | A1 |
| (d) | Translated normal: \(y - 5 = -2(x - k - 4)\) | M1 |
| (d) | Passes through M (9, 6) so \(6 - 5 = -2(9 - k - 4)\) | m1 |
| (d) | \(k = 5.5\) | A1 |
| ALTn 1 | On normal, when \(y=6, 6 - 5 = -2(x - 4)\) | (M1) |
| \(x = 3.5; \quad 3.5 + k = x_M = 9\) | (m1) | (c's \(x_M\)) + k = (c's \(x_M\)) provided c's \(x_M\) is > 0 |
| \(k = 5.5\) | (A1) | (3) |
| ALTn 2 | Line through M parallel to normal at P has equation \(y - 6 = -2(x - 9)\) | (M1) |
| eg Using y=0, for normal at P, (6.5, 0) and for parallel line through M, (12,0) (\(k =\) 12 − 6.5) | (m1) | For any single value of y, finding the x coord of the point on both normal at P and this parallel line through M and then subtracting these x coords in correct order |
| \(k = 5.5\) | (A1) | (3) |
| Total | 11 |
(a) | $\frac{dy}{dx} = \frac{6}{2} x^{-0.5} - 1 = 3x^{-0.5} - 1$ | B2,1 | 2 marks | ACF. If not B2, award B1 for correct differentiation of either $6x^{1/2}$ or $-x - 3$
(b) | $3x^{-0.5} - 1 = 0$ | M1 | Evidence of c's $\frac{dy}{dx}$ equated to 0 to form an equation in x.
(b) | $3x^{-0.5} = 1, \quad x = 9$ | A1F | Only if c's $\frac{dy}{dx} = ax^{-0.5} - 1$ ie $s_M = a^2$
(b) | (y-coordinate of M is) 6 | A1 | 3 marks | NMS scores 0/3
(c) | At P(4,5) $\frac{dy}{dx} = 3(4)^{-0.5} - 1 (=0.5)$ | M1 | Attempt to find c's $\frac{dy}{dx}$ when $x = 4$.
(c) | Gradient of normal = $-2$ | m1 | $m \times m' = -1$ used
(c) | Eqn of normal: $y - 5 = -2(x - 4)$ | A1 | 3 marks | ACF eg $y + 2x = 13$
(d) | Translated normal: $y - 5 = -2(x - k - 4)$ | M1 | Either $x \to x - k$ or $x \to x + k$ with no change to y in cand's eqn of normal seen or used
(d) | Passes through M (9, 6) so $6 - 5 = -2(9 - k - 4)$ | m1 | Subst of c's M coordinates (both +' ve) into cand's eqn of normal with $x \to x - k$ and no change to y in cand's eqn of normal
(d) | $k = 5.5$ | A1 | 3 marks | A correct value of k with no errors seen.
| **ALTn 1** | On normal, when $y=6, 6 - 5 = -2(x - 4)$ | (M1) | Sub answer (b) in answer (c); ie attempt to find $x_M$, the x-coordinate of point on c's normal in part (c) with same y-coord as c's part (b) answer.
| | $x = 3.5; \quad 3.5 + k = x_M = 9$ | (m1) | (c's $x_M$) + k = (c's $x_M$) provided c's $x_M$ is > 0
| | $k = 5.5$ | (A1) | (3) | A correct value of k with no errors seen.
| **ALTn 2** | Line through M parallel to normal at P has equation $y - 6 = -2(x - 9)$ | (M1) | Correct ft eqn using c's M coords, both>0 and c's normal at P
| | eg Using y=0, for normal at P, (6.5, 0) and for parallel line through M, (12,0) ($k =$ 12 − 6.5) | (m1) | For any single value of y, finding the x coord of the point on both normal at P and this parallel line through M and then subtracting these x coords in correct order
| | $k = 5.5$ | (A1) | (3) | A correct value of k with no errors seen.
| **Total** | **11** |
---3 The diagram shows a curve with a maximum point $M$.\\
\includegraphics[max width=\textwidth, alt={}, center]{e183578a-29a8-4112-b941-06c8894ed078-06_512_867_354_589}
The curve is defined for $x > 0$ by the equation
$$y = 6 x ^ { \frac { 1 } { 2 } } - x - 3$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence find the $y$-coordinate of the maximum point $M$.
\item Find an equation of the normal to the curve at the point $P ( 4,5 )$.
\item It is given that the normal to the curve at $P$, when translated by the vector $\left[ \begin{array} { l } k \\ 0 \end{array} \right]$, passes through the point $M$. Find the value of the constant $k$.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2016 Q3 [11]}}