| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find quotient and remainder by division |
| Difficulty | Standard +0.3 This is a standard C2 polynomial question with routine algebraic division, factorization, and integration. Part (i) is straightforward polynomial division, (ii) uses the result to find roots by factoring, (iii) is simple differentiation to find stationary points, and (iv) requires definite integration between roots. All techniques are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07i Differentiate x^n: for rational n and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Q = x^2 - 4x + 3\) | M1 | Attempt complete division by \((x+1)\); must obtain at least the quotient with all 3 terms attempted |
| A1 | Obtain fully correct quotient | |
| \(R = 0\) | A1 | Obtain remainder as 0, must be stated explicitly; allow 'no remainder' for 'remainder \(= 0\)'; \(f(-1) = 0\) not sufficient unless identified as remainder [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 - 4x + 3 = (x-1)(x-3)\) | M1 | Attempt to solve their quadratic quotient |
| \(x = 1, 3\) | A1 | Obtain \(x = 1, 3\); M1A1 if both roots just stated with no method |
| \(x = -1\) | B1 [3] | State \(x = -1\); independent of M mark; B0 if \(x = -1\) is clearly result of solving quadratic quotient only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 4x^3 - 12x^2 - 4x + 12\) | M1 | Attempt differentiation; decrease in power by 1 for at least 3 terms |
| \(4x^3 - 12x^2 - 4x + 12 = 0\) hence \(x^3 - 3x^2 - x + 3 = 0\) AG | A1 [2] | Equate to 0 and rearrange to given answer; must equate to 0 before dividing by 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\frac{1}{5}x^5 - x^4 - \frac{2}{3}x^3 + 6x^2 + 9x\right]_{-1}^{3}\) | M1* | Attempt integration — increase in power by 1 for at least 3 terms; must be integrating equation of curve, not \(f(x)\) |
| \(= \left(\frac{153}{5}\right) - \left(-\frac{23}{15}\right)\) | A1 | Obtain fully correct expression — allow unsimplified coefficients; allow presence of \(+c\) |
| Obtain \(\frac{512}{15}\) | M1d* | Attempt correct use of correct limits — must be \(F(3) - F(-1)\); correct order and subtraction; could find area between 1 and 3 but must double for M1 |
| A1 | Obtain \(\frac{512}{15}\), or any exact equiv — decimal equiv must be exact i.e. \(34.1\overline{3}\); A0 for 34.13, 34.133... |
# Question 7:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Q = x^2 - 4x + 3$ | M1 | Attempt complete division by $(x+1)$; must obtain at least the quotient with all 3 terms attempted |
| | A1 | Obtain fully correct quotient |
| $R = 0$ | A1 | Obtain remainder as 0, must be stated explicitly; allow 'no remainder' for 'remainder $= 0$'; $f(-1) = 0$ not sufficient unless identified as remainder [3] |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - 4x + 3 = (x-1)(x-3)$ | M1 | Attempt to solve their quadratic quotient |
| $x = 1, 3$ | A1 | Obtain $x = 1, 3$; M1A1 if both roots just stated with no method |
| $x = -1$ | B1 [3] | State $x = -1$; independent of M mark; B0 if $x = -1$ is clearly result of solving quadratic quotient only |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 4x^3 - 12x^2 - 4x + 12$ | M1 | Attempt differentiation; decrease in power by 1 for at least 3 terms |
| $4x^3 - 12x^2 - 4x + 12 = 0$ hence $x^3 - 3x^2 - x + 3 = 0$ **AG** | A1 [2] | Equate to 0 and rearrange to given answer; must equate to 0 before dividing by 4 |
## Question 7 (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{1}{5}x^5 - x^4 - \frac{2}{3}x^3 + 6x^2 + 9x\right]_{-1}^{3}$ | M1* | Attempt integration — increase in power by 1 for at least 3 terms; must be integrating equation of curve, not $f(x)$ |
| $= \left(\frac{153}{5}\right) - \left(-\frac{23}{15}\right)$ | A1 | Obtain fully correct expression — allow unsimplified coefficients; allow presence of $+c$ |
| Obtain $\frac{512}{15}$ | M1d* | Attempt correct use of correct limits — must be $F(3) - F(-1)$; correct order and subtraction; could find area between 1 and 3 but must double for M1 |
| | A1 | Obtain $\frac{512}{15}$, or any exact equiv — decimal equiv must be exact i.e. $34.1\overline{3}$; A0 for 34.13, 34.133... |
---7 The cubic polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = x ^ { 3 } - 3 x ^ { 2 } - x + 3$.\\
(i) Find the quotient and remainder when $\mathrm { f } ( x )$ is divided by $( x + 1 )$.\\
(ii) Hence find the three roots of the equation $\mathrm { f } ( x ) = 0$.\\
\includegraphics[max width=\textwidth, alt={}, center]{555f7205-5e2a-4471-901d-d8abc9dd4b4a-3_540_718_1466_660}
The diagram shows the curve $C$ with equation $y = x ^ { 4 } - 4 x ^ { 3 } - 2 x ^ { 2 } + 12 x + 9$.\\
(iii) Show that the $x$-coordinates of the stationary points on $C$ are given by $x ^ { 3 } - 3 x ^ { 2 } - x + 3 = 0$.\\
(iv) Use integration to find the exact area of the region enclosed by $C$ and the $x$-axis.
\hfill \mbox{\textit{OCR C2 2016 Q7 [12]}}