| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Sketching Polynomial Curves |
| Difficulty | Moderate -0.3 This is a structured multi-part question requiring polynomial expansion verification, discriminant analysis, solving a cubic equation using a given root, and curve sketching. While it involves multiple techniques, each step is guided and uses standard C1 methods (factorization, quadratic formula, basic sketching). The question is slightly easier than average because the factorization is provided rather than requiring students to find it independently. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Expand \((x-3)(2x^2+5x+4)\): | M1 | Attempt expansion |
| \(= 2x^3+5x^2+4x-6x^2-15x-12\) | A1 | Correct expansion |
| \(= 2x^3 - x^2 - 11x - 12\) | A1 | Shown correctly |
| Discriminant of \(2x^2+5x+4\): \(25 - 32 = -7 < 0\) | M1 A1 | Hence only one real root \(x=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(2) = 16 - 4 - 22 - 12 = -22\) ✓ | B1 | Verification shown |
| \(f(x) + 22 = 2x^3 - x^2 - 11x + 10\) | M1 | Form new equation |
| \((x-2)\) is factor; divide: \((x-2)(2x^2+3x-5)\) | M1 A1 | Correct quotient |
| \((x-2)(2x+5)(x-1) = 0\) | M1 | Factorise quadratic |
| \(x = -\frac{5}{2}\) or \(x = 1\) | A1 | Both other roots correct |
| Answer | Marks | Guidance |
|---|---|---|
| Cubic curve correct shape | B1 | Positive cubic |
| Crosses \(x\)-axis at \(x = 3\) only for \(y = f(x)\) | B1 | \(x=3\) labelled, no other crossing |
| \(y\)-intercept at \((0, -12)\) | B1 | Correct \(y\)-intercept |
## Question 13:
**(i)**
Expand $(x-3)(2x^2+5x+4)$: | M1 | Attempt expansion
$= 2x^3+5x^2+4x-6x^2-15x-12$ | A1 | Correct expansion
$= 2x^3 - x^2 - 11x - 12$ | A1 | Shown correctly
Discriminant of $2x^2+5x+4$: $25 - 32 = -7 < 0$ | M1 A1 | Hence only one real root $x=3$
**(ii)**
$f(2) = 16 - 4 - 22 - 12 = -22$ ✓ | B1 | Verification shown
$f(x) + 22 = 2x^3 - x^2 - 11x + 10$ | M1 | Form new equation
$(x-2)$ is factor; divide: $(x-2)(2x^2+3x-5)$ | M1 A1 | Correct quotient
$(x-2)(2x+5)(x-1) = 0$ | M1 | Factorise quadratic
$x = -\frac{5}{2}$ or $x = 1$ | A1 | Both other roots correct
**(iii)**
Cubic curve correct shape | B1 | Positive cubic
Crosses $x$-axis at $x = 3$ only for $y = f(x)$ | B1 | $x=3$ labelled, no other crossing
$y$-intercept at $(0, -12)$ | B1 | Correct $y$-intercept13 A cubic polynomial is given by $\mathrm { f } ( x ) = 2 x ^ { 3 } - x ^ { 2 } - 11 x - 12$.\\
(i) Show that $( x - 3 ) \left( 2 x ^ { 2 } + 5 x + 4 \right) = 2 x ^ { 3 } - x ^ { 2 } - 11 x - 12$.
Hence show that $\mathrm { f } ( x ) = 0$ has exactly one real root.\\
(ii) Show that $x = 2$ is a root of the equation $\mathrm { f } ( x ) = - 22$ and find the other roots of this equation.\\
(iii) Using the results from the previous parts, sketch the graph of $y = \mathrm { f } ( x )$.
\hfill \mbox{\textit{OCR MEI C1 2007 Q13 [12]}}