| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Polynomial division before integration |
| Difficulty | Standard +0.3 This is a standard two-part question requiring polynomial long division followed by term-by-term integration. Part (i) is routine algebraic manipulation with the answer provided for verification. Part (ii) involves integrating a polynomial plus a rational function (x/(x²+3)), which is a textbook application of reverse chain rule. The question is slightly above average difficulty due to the multi-step nature and need to recognize the reverse chain rule, but remains a standard C4 exercise with no novel problem-solving required. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Long division: for leading term \(3x\) in quotient | B1 | Words quotient and remainder need not be explicit |
| Sufficient evidence of div process (\(3x\), mult back, attempt sub) | M1 | |
| (Quotient) \(= 3x-1\) | A1 | |
| (Remainder) \(= x\) AG | A1 4 | No wrong working, partic on penult line |
| Identity: \(3x^3-x^2+10x-3 = Q(x^2+3)+R\) | *M1 | |
| \(Q=ax+b\), \(R=cx+d\) & attempt at least 2 operations | dep*M1 | If \(a=3\), this \(\Rightarrow\) 1 operation |
| \(a=3, b=-1\) | A1 | |
| \(c=1, d=0\) | A1 | No wrong working anywhere |
| Inspection: \(3x^3-x^2+10x-3=(x^2+3)(3x-1)+x\) | B2 | or state quotient \(= 3x-1\) |
| Clear demonstration of LHS = RHS | B2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Change integrand to 'their (i) quotient' \(+\frac{x}{x^2+3}\) | M1 | |
| Correct FT integration of 'their (i) quotient' | \(\sqrt{}\)A1 | |
| \(\int\frac{x}{x^2+3}\,dx = \frac{1}{2}\ln(x^2+3)\) | A1 | |
| Exact value of integral \(= \frac{1}{2}+\frac{1}{2}\ln 4-\frac{1}{2}\ln 3\) AEF ISW | A1 4 | Answer as decimal value (only) \(\to\) A0 |
# Question 3:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Long division: for leading term $3x$ in quotient | B1 | Words quotient and remainder need not be explicit |
| Sufficient evidence of div process ($3x$, mult back, attempt sub) | M1 | |
| (Quotient) $= 3x-1$ | A1 | |
| (Remainder) $= x$ **AG** | A1 **4** | No wrong working, partic on penult line |
| Identity: $3x^3-x^2+10x-3 = Q(x^2+3)+R$ | *M1 | |
| $Q=ax+b$, $R=cx+d$ & attempt at least 2 operations | dep*M1 | If $a=3$, this $\Rightarrow$ 1 operation |
| $a=3, b=-1$ | A1 | |
| $c=1, d=0$ | A1 | No wrong working anywhere |
| Inspection: $3x^3-x^2+10x-3=(x^2+3)(3x-1)+x$ | B2 | or state quotient $= 3x-1$ |
| Clear demonstration of LHS = RHS | B2 | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Change integrand to 'their (i) quotient' $+\frac{x}{x^2+3}$ | M1 | |
| Correct FT integration of 'their (i) quotient' | $\sqrt{}$A1 | |
| $\int\frac{x}{x^2+3}\,dx = \frac{1}{2}\ln(x^2+3)$ | A1 | |
| Exact value of integral $= \frac{1}{2}+\frac{1}{2}\ln 4-\frac{1}{2}\ln 3$ AEF ISW | A1 **4** | Answer as decimal value (only) $\to$ A0 |
---3 (i) Find the quotient when $3 x ^ { 3 } - x ^ { 2 } + 10 x - 3$ is divided by $x ^ { 2 } + 3$, and show that the remainder is $x$.\\
(ii) Hence find the exact value of
$$\int _ { 0 } ^ { 1 } \frac { 3 x ^ { 3 } - x ^ { 2 } + 10 x - 3 } { x ^ { 2 } + 3 } \mathrm {~d} x$$
\hfill \mbox{\textit{OCR C4 2011 Q3 [8]}}