Edexcel A-Level Mathematics A2 P4.2 Algebra and Functions QuestionsPractise rewriting rational functions as partial fractions and solving for constants in forms used for later integration.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsexpress rational functions in the requested partial fraction form with constants A, B and Ccompare coefficients or substitute roots to find constants in linear and repeated factors
Question 1[Maximum number: 3]Figure 2Figure 2 shows a sketch of the curve defined by the parametric equationsx=t2+2ty=2t(3−t)a⩽t⩽bx=t^{2}+2 t \quad y=\frac{2}{t(3-t)} \quad a \leqslant t \leqslant bx=t2+2ty=t(3−t)2a⩽t⩽bwhere a and b are constants.The ends of the curve lie on the line with equation y=1Write t+1t(3−t)\frac{t+1}{t(3-t)}t(3−t)t+1 in partial fractions.Mark as masteredShow Answer(ii)t+1t(3−t)=At+B3−t\frac{t+1}{t(3-t)}=\frac{A}{t}+\frac{B}{3-t}t(3−t)t+1=tA+3−tB⇒t+1=A(3−t)+Bt\Rightarrow t+1=A(3-t)+B t⇒t+1=A(3−t)+BtE.g. t=0⇒1=3A⇒A=13;t=3⇒4=3B⇒B=43t=0 \Rightarrow 1=3 A \Rightarrow A=\frac{1}{3} ; t=3 \Rightarrow 4=3 B \Rightarrow B=\frac{4}{3}t=0⇒1=3A⇒A=31;t=3⇒4=3B⇒B=34t+1t(3−t)=13t+43(3−t)\frac{t+1}{t(3-t)}=\frac{1}{3 t}+\frac{4}{3(3-t)}t(3−t)t+1=3t1+3(3−t)4Add to Test
Question 2[Maximum number: 3]f(x)=5x+10(1−x)(2+3x)f(x)=\frac{5 x+10}{(1-x)(2+3 x)}f(x)=(1−x)(2+3x)5x+10Write f(x) in partial fraction form.Mark as masteredShow Answer5x+10(1−x)(2+3x)≡A1−x+B2+3x⇒\frac{5 x+10}{(1-x)(2+3 x)} \equiv \frac{A}{1-x}+\frac{B}{2+3 x} \Rightarrow(1−x)(2+3x)5x+10≡1−xA+2+3xB⇒ Value for A or BOne correct value, either A=3 or B=4Correct PF form 31−x+42+3x\frac{3}{1-x}+\frac{4}{2+3 x}1−x3+2+3x4(3)Add to Test
Question 3[Maximum number: 4]f(x)=2x4+15x3+35x2+21x−4(x+3)2x∈Rx>−3f(x)=\frac{2 x^{4}+15 x^{3}+35 x^{2}+21 x-4}{(x+3)^{2}} \quad x \in \mathbb{R} \quad x>-3f(x)=(x+3)22x4+15x3+35x2+21x−4x∈Rx>−3Find the values of the constants A, B, C and D such thatf(x)=Ax2+Bx+C+D(x+3)2\mathrm{f}(x)=A x^{2}+B x+C+\frac{D}{(x+3)^{2}}f(x)=Ax2+Bx+C+(x+3)2DMark as masteredShow AnswerAny correct constant, so A=2, B=3, C=-1, or D=52x4+15x3+35x2+21x−4=Ax2(x+3)2+Bx(x+3)2+C(x+3)2+D2x^4+15x^3+35x^2+21x-4=Ax^2(x+3)^2+Bx(x+3)^2+C(x+3)^2+D2x4+15x3+35x2+21x−4=Ax2(x+3)2+Bx(x+3)2+C(x+3)2+D⇒A=…, B=…, C=…, D=…\Rightarrow A=\ldots,\ B=\ldots,\ C=\ldots,\ D=\ldots⇒A=…, B=…, C=…, D=…or2x4+15x3+35x2+21x−4÷(x2+6x+9)=…x2+…x+…+…(x+3)22x^4+15x^3+35x^2+21x-4\div(x^2+6x+9)=\ldots x^2+\ldots x+\ldots+\frac{\ldots}{(x+3)^2}2x4+15x3+35x2+21x−4÷(x2+6x+9)=…x2+…x+…+(x+3)2…2 correct of A=2, B=3, C=-1, D=5A=2, B=3, C=-1, D=5(4)Add to Test