OCR Further Pure Core 2 2018 March — Question 8 12 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2018
SessionMarch
Marks12
TopicIntegration with Partial Fractions
TypeFactorization then partial fractions
DifficultyChallenging +1.8 This is a challenging Further Maths integration problem requiring partial fractions with complex roots, leading to an inverse tangent integral. The 12-mark allocation and 'show detailed reasoning' instruction indicate substantial work. Students must factor the cubic denominator (likely grouping), decompose into partial fractions including a quadratic factor, integrate to obtain both logarithmic and arctangent terms, then carefully evaluate limits to reach the specific answer involving π and ln. While systematic, it requires multiple advanced techniques and careful algebraic manipulation throughout.
Spec1.08j Integration using partial fractions4.05c Partial fractions: extended to quadratic denominators
In this question you must show detailed reasoning. Show that \(\int_0^2 \frac{2x^2 + 3x - 1}{x^3 - 3x^2 + 4x - 12} dx = \frac{3}{8}\pi - \ln 9\). [12]
AnswerMarks Guidance
\(f(x) = x^3 - 3x^2 + 4x - 12\) and \(f(3) = 0\)M1 Attempt to find a factor of the denominator
\(x - 3\) is a factorA1 [2.2a]
\(f(x) = (x - 3)(x^2 + 4)\)A1 [2.2a]
\(\frac{2x^2 + 3x - 1}{x^3 - 3x^2 + 4x - 12} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 4}\)B1 Correct partial fraction expression using their factorised cubic
\(2x^2 + 3x - 1 = A(x^2 + 4) + (Bx + C)(x - 3)\)M1 A correct method for finding A, B or C (no terms omitted)
\(A = 2\)A1
\(C = 3\)A1
\(B = 0\)A1
\(\int \frac{1}{x-3}dx = \lnx - 3 \)
\(\int \frac{1}{x^2+4}dx = \frac{1}{2}\tan^{-1}\frac{x}{2}\)B1 Ignore additive constant. Allow for their (numerical) a.
\(2\ln2-3 + \frac{3}{2}\tan^{-1}\frac{2}{2} - (2\ln
\(-2\ln 3 + \frac{3}{2} \cdot \frac{\pi}{4} - \frac{3}{8}\pi - \ln 9\)A1 AG Must see evidence of use of laws of logs.
[12]
$f(x) = x^3 - 3x^2 + 4x - 12$ and $f(3) = 0$ | M1 | Attempt to find a factor of the denominator | Could be by factorising pairs of terms

$x - 3$ is a factor | A1 [2.2a]
$f(x) = (x - 3)(x^2 + 4)$ | A1 [2.2a]

$\frac{2x^2 + 3x - 1}{x^3 - 3x^2 + 4x - 12} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 4}$ | B1 | Correct partial fraction expression using their factorised cubic

$2x^2 + 3x - 1 = A(x^2 + 4) + (Bx + C)(x - 3)$ | M1 | A correct method for finding A, B or C (no terms omitted)

$A = 2$ | A1
$C = 3$ | A1
$B = 0$ | A1

$\int \frac{1}{x-3}dx = \ln|x - 3|$ | B1 | Ignore multiplicative or additive constant. Condone omitted modulus signs. Allow for (any one of) their linear factor(s).

$\int \frac{1}{x^2+4}dx = \frac{1}{2}\tan^{-1}\frac{x}{2}$ | B1 | Ignore additive constant. Allow for their (numerical) a.

$2\ln|2-3| + \frac{3}{2}\tan^{-1}\frac{2}{2} - (2\ln|-3| + \frac{3}{2}\tan^{-1}\frac{0}{2})$ | M1 | Correct substitution of limits (and subtraction) into an integral containing ln and $\tan^{-1}$. Condone $\tan^{-1}0$ missing.

$-2\ln 3 + \frac{3}{2} \cdot \frac{\pi}{4} - \frac{3}{8}\pi - \ln 9$ | A1 | AG Must see evidence of use of laws of logs.

[12]

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In this question you must show detailed reasoning.

Show that $\int_0^2 \frac{2x^2 + 3x - 1}{x^3 - 3x^2 + 4x - 12} dx = \frac{3}{8}\pi - \ln 9$. [12]

\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q8 [12]}}
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