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This is the second of our many quizzes on Partial fraction decomposition. Click on this to access more such competitive quizzes on Calculus.

Partial fraction decomposition Quiz 2

mathematics

    What is partial fraction of  $2\left( \dfrac{{{x}^{2}}-1}{{{x}^{3}}+{{x}^{2}}+x} \right)$

 Multiply 2 with the terms in numerator, then take $x$ common in the denominator. Now convert the obtained expression as partial fraction and solve.

  $  2\left( \dfrac{{{x}^{2}}-1}{{{x}^{3}}+{{x}^{2}}+x} \right)=\left( \dfrac{2{{x}^{2}}-2}{x\left( {{x}^{2}}+x+1 \right)} \right) \\ $ Writing  in the form of partial fraction, we get $ \\ \Rightarrow \left( \dfrac{2{{x}^{2}}-2}{x\left( {{x}^{2}}+x+1 \right)} \right)=\dfrac{A}{x}+\dfrac{Bx+C}{\left( {{x}^{2}}+x+1 \right)} \\   \Rightarrow 2{{x}^{2}}-2=x\left( Bx+C \right)+A\left( {{x}^{2}}+x+1 \right) \\   \Rightarrow 2{{x}^{2}}-2=A{{x}^{2}}+Ax+A+B{{x}^{2}}+Cx \\  \Rightarrow 2{{x}^{2}}-2={{x}^{2}}\left( A+B \right)+x\left( A+C \right)+A \\ $ Comparing the coefficients of the like terms and solving, we get $ \\  A=-2,B=4,C=2 \\  \therefore 2\left( \dfrac{{{x}^{2}}-1}{{{x}^{3}}+{{x}^{2}}+x} \right)=\dfrac{-2}{x}+\dfrac{4x+2}{\left( {{x}^{2}}+x+1 \right)} $

 Find $\dfrac{x}{\left( x+1 \right)\left( x-4 \right)}$   using partial fraction.

 We use the partial fraction formula $\dfrac{A}{\left( x+a \right)}+\dfrac{B}{\left( x+b \right)}$ to solve the given equation.

  Using partial fraction, we get $ \\ \Rightarrow \dfrac{x}{\left( x+1 \right)\left( x-4 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{B}{\left( x-4 \right)}......(i) \\   \Rightarrow x=A\left( x-4 \right)+B\left( x+1 \right) \\  \Rightarrow x=Ax-4A+Bx+B \\  \Rightarrow x=x\left( A+B \right)-4A+B \\ $ Comparing the coefficients of the like terms and solving, we get $ \\  A=\dfrac{1}{5},B=\dfrac{4}{5} \\ $ Substituting these in equation $(i)$, we get $ \\  \therefore \dfrac{x}{\left( x+1 \right)\left( x-4 \right)}=\dfrac{1}{5\left( x+1 \right)}+\dfrac{4}{5\left( x-4 \right)} $

  Decompose $\dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}$  using partial fraction then find the value for $z=1$.

 Use the partial fraction formula $\dfrac{A}{\left( x+a \right)}+\dfrac{B}{\left( x+b \right)}$ to solve the given equation, then substitute $z=1$ and simplify.

 $  \dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}=\dfrac{1}{{{\left( z-1 \right)}^{2}}\left( z+1 \right)} \\  $ Using partial fraction, we get $ \\   \Rightarrow \dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}=\dfrac{A}{\left( z-1 \right)}+\dfrac{B}{{{\left( z-1 \right)}^{2}}}+\dfrac{C}{\left( z+1 \right)}......(i) \\   \Rightarrow \dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}=\dfrac{A(z-1)\left( z+1 \right)+B(z+1)+C{{\left( z-1 \right)}^{2}}}{{{\left( z-1 \right)}^{2}}\left( z+1 \right)} \\   \Rightarrow 1=A(z-1)\left( z+1 \right)+B(z+1)+C{{\left( z-1 \right)}^{2}} \\   \Rightarrow 1={{z}^{2}}A+{{z}^{2}}C+zB-2zC-A+B+C \\ $ Comparing the coefficients of the like terms and solving, we get $ \\   A=-\dfrac{1}{4},B=\dfrac{1}{2},C=\dfrac{1}{4} \\$ Substituting these in equation $(i)$, we get $ \\   \therefore \dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}=-\dfrac{1}{4\left( z-1 \right)}+\dfrac{1}{2{{\left( z-1 \right)}^{2}}}+\dfrac{1}{4\left( z+1 \right)} \\  \text{When }z=1, \\  \dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}=-\dfrac{1}{4\left( 1-1 \right)}+\dfrac{1}{2{{\left( 1-1 \right)}^{2}}}+\dfrac{1}{4\left( 1+1 \right)} \\  \therefore \dfrac{1}{\left( 1-z \right)\left( 1-{{z}^{2}} \right)}=\text{undefined}$

  What is the constant term after the partial fraction of $\dfrac{{{x}^{6}}}{\left( {{x}^{2}}-4 \right)}$.

  The degree of the numerator is greater than the degree of the denominator, so first do a long division on the numerator and denominator and simplify. Now use the partial fraction formula to solve the given equation and find the constant term.

 Long division of numerator and denominator gives, $ \\   \dfrac{{{x}^{6}}}{\left( {{x}^{2}}-4 \right)}={{x}^{4}}+4{{x}^{2}}+16+\dfrac{64}{\left( {{x}^{2}}-4 \right)}.....(i) \\   \text{Using partial fraction, we get} \\  \Rightarrow \dfrac{64}{\left( {{x}^{2}}-4 \right)}=\dfrac{A}{(x-2)}+\dfrac{B}{\left( x+2 \right)}......(ii) \\   \Rightarrow 64=A(x+2)+B(x-2) \\   \Rightarrow 64=x\left( A+B \right)+2A-2B \\ $ Comparing the coefficients of the like terms and solving, we get $ \\   A=16,B=-16 \\ $ Substituting these in equation $(ii)$, we get $ \\   \therefore \dfrac{64}{\left( {{x}^{2}}-4 \right)}=\dfrac{16}{(x-2)}-\dfrac{16}{\left( x+2 \right)} \\ $ Substituting these in equation $(i)$, we get $  \\   \dfrac{{{x}^{6}}}{\left( {{x}^{2}}-4 \right)}={{x}^{4}}+4{{x}^{2}}+16+\dfrac{16}{(x-2)}-\dfrac{16}{\left( x+2 \right)} $

 The difference between the coefficient of the terms of partial fraction of  <br>$f(x)=$$\dfrac{2(1+x)}{x\left( {{x}^{2}}+3x+2 \right)}$ will be the value of $x$ .Then find the value for that $f(x)$ .

 First find the factors of the denominator and simplify the given expression. Then find the values of the coefficients of the partial fraction, i.e., A and B. Then find the difference, this will be the value of $x$. Now find $f(x)$.

$  \dfrac{2(1+x)}{x\left( {{x}^{2}}+3x+2 \right)}=\dfrac{2(1+x)}{x\left( x+1 \right)\left( x+2 \right)} \\   \dfrac{2(1+x)}{x\left( {{x}^{2}}+3x+2 \right)}=\dfrac{2}{x\left( x+2 \right)} \\ $ Using partial fraction, we get $ \\ \Rightarrow \dfrac{2}{x\left( x+2 \right)}=\dfrac{A}{x}+\dfrac{B}{\left( x+2 \right)}......(ii) \\   \Rightarrow 2=A(x+2)+Bx \\   \Rightarrow 2=x\left( A+B \right)+2A \\ $ Comparing the coefficients of the like terms and solving, we get $ \\  A=1,B=-1 \\ $ So, the difference between the coefficients of the of the terms of partial fraction is, $ \\  A-B=1-(-1)=2 \\   \therefore f(2)=\dfrac{2(1+2)}{2\left( {{2}^{2}}+3(2)+2 \right)} \\  \Rightarrow f(2)=\dfrac{3}{12}=\dfrac{1}{4} $