Pre-U Pre-U 9794/1 2018 June — Question 5 10 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2018
SessionJune
Marks10
TopicAreas by integration
TypeArea between curve and line
DifficultyStandard +0.3 This is a standard area-between-curves question requiring differentiation to find the tangent equation, solving a cubic to verify intersection, and integration to find area. All steps are routine A-level techniques with no novel insight required, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.08f Area between two curves: using integration

5 \includegraphics[max width=\textwidth, alt={}, center]{7895dcbc-2ae0-498f-8770-7b738feed7c9-2_746_1182_1304_479} The diagram shows the curve with equation \(y = x ^ { 3 } + 2 x ^ { 2 } - 13 x + 10\) and the tangent to the curve at the point ( 2,0 ).
  1. Find the equation of this tangent and verify that the tangent intersects the curve when \(x = - 6\).
  2. Calculate the exact area of the region bounded by the curve and the tangent.

Question 5:
(i) \(\frac{\text{d}y}{\text{d}x} = 3x^2 + 4x - 13\) M1 (Differentiate by showing a decrease in power by 1 in at least two terms)
At \(x = 2\), \(m = 7\) M1 (Substitute \(x = 2\) in their derivative and a numerical result)
\(y = 7x - 14\) A1
\(y = 7(-6) - 14 = -56\)
\(y = (-6)^3 + 2(-6)^2 - 13(-6) + 10 = -56\)
OR use \(x^3 + 2x^2 - 20x + 24 = -216 + 72 + 120 + 24 = 0\) A1 (Confirm equality)
(ii) \(x^3 + 2x^2 - 13x + 10 - (7x - 14)\) M1 (Intention shown to subtract curve – tangent, even as two separate integrals)
\(\int_a^b (x^3 + 2x^2 - 20x + 24)\,\text{d}x\) M1 (Integrate one or other cubic expressions)
\(\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_a^b\) A1
\(F(b) - F(a)\) M1 (Evaluate their integral with their \(a\) and \(b\) as \(F(b) - F(a)\) only)
Show use of correct limits \(\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_{-6}^{2}\) A1
\(224\left(=\frac{2688}{12}\right) - \frac{307}{12} + \frac{1728}{12}\left(=144\right) - \frac{13}{12}\)
\(\frac{1024}{3}\) A1 (Accept \(341\frac{1}{3}\) or exact equiv.)
Total: 10 marks
**Question 5:**

**(i)** $\frac{\text{d}y}{\text{d}x} = 3x^2 + 4x - 13$ **M1** (Differentiate by showing a decrease in power by 1 in at least two terms)

At $x = 2$, $m = 7$ **M1** (Substitute $x = 2$ in their derivative and a numerical result)

$y = 7x - 14$ **A1**

$y = 7(-6) - 14 = -56$
$y = (-6)^3 + 2(-6)^2 - 13(-6) + 10 = -56$
OR use $x^3 + 2x^2 - 20x + 24 = -216 + 72 + 120 + 24 = 0$ **A1** (Confirm equality)

**(ii)** $x^3 + 2x^2 - 13x + 10 - (7x - 14)$ **M1** (Intention shown to subtract curve – tangent, even as two separate integrals)

$\int_a^b (x^3 + 2x^2 - 20x + 24)\,\text{d}x$ **M1** (Integrate one or other cubic expressions)

$\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_a^b$ **A1**

$F(b) - F(a)$ **M1** (Evaluate their integral with their $a$ and $b$ as $F(b) - F(a)$ only)

Show use of correct limits $\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_{-6}^{2}$ **A1**

$224\left(=\frac{2688}{12}\right) - \frac{307}{12} + \frac{1728}{12}\left(=144\right) - \frac{13}{12}$

$\frac{1024}{3}$ **A1** (Accept $341\frac{1}{3}$ or exact equiv.)

**Total: 10 marks**
5\\
\includegraphics[max width=\textwidth, alt={}, center]{7895dcbc-2ae0-498f-8770-7b738feed7c9-2_746_1182_1304_479}

The diagram shows the curve with equation $y = x ^ { 3 } + 2 x ^ { 2 } - 13 x + 10$ and the tangent to the curve at the point ( 2,0 ).\\
(i) Find the equation of this tangent and verify that the tangent intersects the curve when $x = - 6$.\\
(ii) Calculate the exact area of the region bounded by the curve and the tangent.

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2018 Q5 [10]}}