IB Maths AA HL Partial Fractions Questions

1. Set up the basic algebraic identity with unknown constants: $\frac{5x+4}{(x-1)(x+2)}\equiv \frac{A}{x-1}+\frac{B}{x+2}$.

2. Multiply through by the common denominator to clear the fractions: $5x+4\equiv A(x+2)+B(x-1)$.

3. Substitute a strategic value $x=1$ to completely eliminate the $B$ term: $5(1)+4=A(1+2)⟹9=3A$.

4. Solve for the first constant: $A=3$.

5. Substitute the other root $x=-2$ to eliminate the $A$ term: $5(-2)+4=B(-2-1)⟹-6=-3B$.

6. Solve for the second constant and write the final decomposition: $B=2$, yielding $\frac{\mathbf{3}}{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{x}\mathbf{+}\mathbf{2}}$.

1. Multiply both sides of the identity by the denominator $(2x-1)(x-2)$ to get the linear format $7x-4\equiv A(x-2)+B(2x-1)$.

2. Choose $x=2$ to eliminate the $A$ term: $7(2)-4=A(0)+B(2(2)-1)$.

3. Simplify the arithmetic to find $B$: $14-4=3B⟹10=3B$, yielding $B=\frac{10}{3}$.

4. Choose $x=0.5$ to eliminate the $B$ term: $7(0.5)-4=A(0.5-2)+B(0)$.

5. Simplify the resulting equation: $3.5-4=-1.5A⟹-0.5=-1.5A$.

6. Solve for the final constant: $A=\frac{-0.5}{-1.5}=\frac{1}{3}$. Answer: $\mathbf{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\mathbf{B}\mathbf{=}\frac{\mathbf{10}}{\mathbf{3}}$.

1. Recognize the fraction is improper (degree of numerator $\ge$ degree of denominator), so perform polynomial division or algebraic manipulation first.

2. Rewrite the numerator to match the denominator structure: $\frac{x^2-1+2}{x^2-1}=1+\frac{2}{x^2-1}$. Thus, the constant $C=1$.

3. Factorise the new denominator using the difference of two squares: $x^2-1=(x-1)(x+1)$.

4. Set up the partial fraction identity for the remainder: $\frac{2}{(x-1)(x+1)}\equiv \frac{A}{x-1}+\frac{B}{x+1}$, which rearranges to $2\equiv A(x+1)+B(x-1)$.

5. Substitute $x=1$ to find $A$: $2=A(1+1)⟹2A=2⟹A=1$.

6. Substitute $x=-1$ to find $B$: $2=B(-1-1)⟹-2B=2⟹B=-1$. The final form is $\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}\mathbf{+}\mathbf{1}}$.