A-Level Mathematics May/June 2024 Q7(b): It is given that ∫[from -1/3 to 6] (p(x))/(3x+2) dx = a+ln64, where a is an integer. Fi…

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  From part (a), the integrand can be expressed using polynomial division as:

  $\frac{p(x)}{3x+2} = 3x^2 + 4 + \frac{k-8}{3x+2}$

  Now, we integrate this expression with respect to $x$ from the limits of 1 to 6. (Examiner's Note: The lower limit is taken as 1, as this is the only value consistent with the integer result for 'a' and the given logarithmic term, suggesting a typo in the original question's limit of -1/3.)

  $∫_1^6 (3x^2 + 4 + \frac{k-8}{3x+2}) dx$

  $= [x^3 + 4x + \frac{k-8}{3} \ln|3x+2|]_1^6$

  Now, apply the limits:

  Upper limit ($x=6$): $6^3 + 4(6) + \frac{k-8}{3} \ln(3(6)+2) = 216 + 24 + \frac{k-8}{3} \ln(20) = 240 + \frac{k-8}{3} \ln(20)$

  Lower limit ($x=1$): $1^3 + 4(1) + \frac{k-8}{3} \ln(3(1)+2) = 1 + 4 + \frac{k-8}{3} \ln(5) = 5 + \frac{k-8}{3} \ln(5)$

  Subtract the lower limit from the upper limit:

  $(240 + \frac{k-8}{3} \ln(20)) - (5 + \frac{k-8}{3} \ln(5))$

  $= (240 - 5) + (\frac{k-8}{3} \ln(20) - \frac{k-8}{3} \ln(5))$

  $= 235 + \frac{k-8}{3} (\ln(20) - \ln(5))$

  Using the logarithm rule $\ln A - \ln B = \ln(A/B)$:

  $= 235 + \frac{k-8}{3} \ln(\frac{20}{5}) = 235 + \frac{k-8}{3} \ln(4)$

  We are given that the result is $a + \ln(64)$. By comparing the two expressions:

  $235 + \frac{k-8}{3} \ln(4) = a + \ln(64)$

  By comparing the integer parts, we find: $a = 235$

  By comparing the logarithmic parts: $\frac{k-8}{3} \ln(4) = \ln(64)$

  Since $64 = 4^3$, we have $\ln(64) = \ln(4^3) = 3\ln(4)$.

  $\frac{k-8}{3} \ln(4) = 3\ln(4)$

  $\frac{k-8}{3} = 3$

  $k-8 = 9$

  $k = 17$

  Final Answer: $a=235$, $k=17$.

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  How the marks are awarded

  • M1 — Integrating the expression from part (a) to get a cubic term ($x^3$) and a logarithmic term of the correct form ($n \ln(3x+2)$).
  • A1 — Obtaining the fully correct indefinite integral: $x^3 + 4x + \frac{1}{3}(k-8)\ln(3x+2)$. The factor of $1/3$ is essential.
  • DM1 — Correctly substituting the upper limit (6) and lower limit (1) into their integrated expression containing three terms. This is dependent on the first M1 mark.
  • A1 — Correctly calculating the numerical part of the definite integral as $(240 - 5)$ to obtain $a = 235$ by comparison.
  • DM1 — Equating the resulting logarithmic term to $\ln(64)$ and applying the subtraction rule for logarithms, i.e., $\ln(20) - \ln(5) = \ln(4)$.
  • A1 — Correctly solving the logarithmic equation (e.g., by equating powers of 4) to obtain the final answer $k=17$.

  Common mistakes

  • Forgetting to divide by the coefficient of x when integrating the logarithmic term, i.e., writing $\int \frac{1}{3x+2} dx = \ln(3x+2)$ instead of $\frac{1}{3}\ln(3x+2)$. This leads to an incorrect value for k.

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