A-Level Mathematics October/November 2024 Q7(a): Let f(x) = (5x² +8x+5)/((1+2x)(2+x²)). Express f(x) in partial fractions.

1 answer

• ▲0
  accepted ✓

  Let the expression be written in the form of partial fractions.

  $f(x) = \frac{5x^2 + 8x + 5}{(1+2x)(2+x^2)} \equiv \frac{A}{1+2x} + \frac{Bx+C}{2+x^2}$

  To find the constants A, B, and C, we multiply both sides by the denominator $(1+2x)(2+x^2)$ to get:

  $5x^2 + 8x + 5 \equiv A(2+x^2) + (Bx+C)(1+2x)$

  We can find the value of A by substituting a value of x that makes the term (1+2x) equal to zero. Let $1+2x = 0$, which means $x = -\frac{1}{2}$.

  $5(-\frac{1}{2})^2 + 8(-\frac{1}{2}) + 5 = A(2+(-\frac{1}{2})^2) + (B(-\frac{1}{2})+C)(0)$ $5(\frac{1}{4}) - 4 + 5 = A(2+\frac{1}{4})$ $\frac{5}{4} + 1 = A(\frac{9}{4})$ $\frac{9}{4} = A(\frac{9}{4})$ $A = 1$

  Now we can find B and C by equating coefficients. Expanding the identity: $5x^2 + 8x + 5 \equiv A(2+x^2) + Bx + 2Bx^2 + C + 2Cx$ $5x^2 + 8x + 5 \equiv (A+2B)x^2 + (B+2C)x + (2A+C)$

  Equating the coefficients of $x^2$: $5 = A+2B$ Since $A=1$, we have: $5 = 1+2B$ $4 = 2B$ $B = 2$

  Equating the constant terms: $5 = 2A+C$ Since $A=1$, we have: $5 = 2(1)+C$ $5 = 2+C$ $C = 3$

  (As a check, we can equate the coefficients of x: $8 = B+2C$. With our values, 2+2(3) = 2+6 = 8, which is correct.)

  Therefore, the expression in partial fractions is:

  $f(x) = \frac{1}{1+2x} + \frac{2x+3}{2+x^2}$

  ---

  How the marks are awarded

  • B1 — Stating the correct form of the partial fractions as $\frac{A}{1+2x} + \frac{Bx+C}{2+x^2}$.
  • M1 — Using a correct method to find a constant. This is shown by creating the identity $5x^2 + 8x + 5 \equiv A(2+x^2) + (Bx+C)(1+2x)$ and then either substituting $x = -1/2$ or starting to equate coefficients.
  • A1 — Correctly obtaining the first constant, $A=1$.
  • A1 — Correctly obtaining the second constant, $B=2$.
  • A1 — Correctly obtaining the third constant, $C=3$, and stating the final answer correctly.

  Common mistakes

  • Using an incorrect form for the irreducible quadratic factor, such as $\frac{B}{2+x^2}$ instead of $\frac{Bx+C}{2+x^2}$. This is a fundamental error that prevents any further marks for the constants.
  • Making an arithmetic error when substituting $x = -1/2$, particularly with the fractions and signs, leading to an incorrect value for A and subsequent constants.
  • Incorrectly expanding the brackets when setting up the equations for equating coefficients, for example getting $(Bx+C)(1+2x) = Bx+C+2Bx^2$ and forgetting the 2Cx term.
  • Solving the simultaneous equations incorrectly after equating coefficients, even if the setup is correct.

  Examiner tip: Always identify the type of each factor in the denominator (linear, repeated linear, or irreducible quadratic) to select the correct form for the partial fractions.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers✓ Official

Your answer

Sign in to answer this question.