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 problem requiring differentiation to find the tangent equation, solving a cubic to verify intersection, and integration of a polynomial. All techniques are routine for A-level; the cubic factorization is given by the verification step, making this 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]{69214874-18a7-495d-892d-2a0a7019cbe9-2_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)
- \(\frac{dy}{dx} = 3x^2 + 4x - 13\) 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)
- \(\int_a^b (x^3 + 2x^2 - 20x + 24)\,dx\) OR \(\int_a^b (x^3 + 2x^2 - 13x + 10)\,dx\) OR \(\int_a^b (x^3 + 2x^2 - 13x + 66)\,dx\) M1 (Integrate one or other cubic expressions)
- \(\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_a^b\) OR \(\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - \frac{13}{2}x^2 + 10x\right]_a^b\) 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 \(\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_{-6}^{2}\); \(224\left(=\frac{2688}{12}\right) - \frac{307}{12} + \frac{1728}{12}\left(=144\right) - \frac{13}{12}\) A1
- \(\frac{1024}{3}\) A1 (Accept \(341\frac{1}{3}\) or exact equiv.)
Total: 10 marks
**Question 5:**

**(i)**
- $\frac{dy}{dx} = 3x^2 + 4x - 13$ **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)
- $\int_a^b (x^3 + 2x^2 - 20x + 24)\,dx$ OR $\int_a^b (x^3 + 2x^2 - 13x + 10)\,dx$ OR $\int_a^b (x^3 + 2x^2 - 13x + 66)\,dx$ **M1** (Integrate one or other cubic expressions)
- $\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_a^b$ OR $\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - \frac{13}{2}x^2 + 10x\right]_a^b$ **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 $\left[\frac{1}{4}x^4 + \frac{2}{3}x^3 - 10x^2 + 24x\right]_{-6}^{2}$; $224\left(=\frac{2688}{12}\right) - \frac{307}{12} + \frac{1728}{12}\left(=144\right) - \frac{13}{12}$ **A1**
- $\frac{1024}{3}$ **A1** (Accept $341\frac{1}{3}$ or exact equiv.)

**Total: 10 marks**
5\\
\includegraphics[max width=\textwidth, alt={}, center]{69214874-18a7-495d-892d-2a0a7019cbe9-2_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]}}
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