Skip to main contentSkip to solution

Match the rational functions with the correct form of the partial fractions.

[ p,q,r,a,b,c,A,B,Cp,q,r,a,b,c,A,B,C are arbitrary constants]

Match List-I with List-II

List-I (Rational Function)List-II (Partial Fraction)
(A) px2+qx+r(x−a)(x2+bx+c)\dfrac{px^2+qx+r}{(x-a)(x^2+bx+c)}(I) Ax−a+B(x−a)2+Cx−b\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}
(B) px2+qx+r(x−a)2(x−b),a≠b\dfrac{px^2+qx+r}{(x-a)^2(x-b)}, a\neq b(II) Ax−a+B(x−a)2\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}
(C) px2+qx+r(x−a)(x−b)(x−c),a≠b≠c\dfrac{px^2+qx+r}{(x-a)(x-b)(x-c)}, a\neq b\neq c(III) Ax−a+Bx+Cx2+bx+c\dfrac{A}{x-a}+\dfrac{Bx+C}{x^2+bx+c}
(D) px+q(x−a)2\dfrac{px+q}{(x-a)^2}(IV) Ax−a+Bx−b+Cx−c\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

Each distinct linear factor contributes a constant numerator, a repeated factor (x−a)2(x-a)^2 contributes two terms, and a non-factorisable quadratic needs a linear numerator Bx+CBx+C. Matching these shapes gives (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question