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JEE Previous Year Question

$\frac{x ^{4}}{( x ^{2} + 1 ) ( x ^{2} + 3 )} =$

Topic: JEE PYQ

Answer: D. Full step-by-step solution below.

A $\frac{Ax + B}{x ^{2} + 1} + \frac{Cx + D}{x ^{2} + 3} for some A, B, C, D \in R ∖{0}$
B $\frac{Ax + B}{x ^{2} + 1} + \frac{Cx}{x ^{2} + 1} for some A, B, C \in R ∖{0}$
C $\frac{Ax}{x ^{2} + 1} + \frac{Bx}{x ^{2} + 3} for some A,B \in R ∖{0}$
D $1 + \frac{Ax + B}{x ^{2} + 1} + \frac{Cx + D}{x ^{2} + 3} for some A, B, C, D \in R$

Correct answer: D

Solution

(d) Given \frac{x ^{4}}{( x ^{2} + 1 ) ( x ^{2} + 3 )} = \frac{x ^{4} + 4 x ^{2} + 3}{x ^{4} + 4 x ^{2} + 3} - \frac{( 4 x ^{2} + 3 )}{( x ^{4} + 4 x ^{2} + 3 )} = 11 - \frac{( 4 x ^{2} + 3 )}{( x ^{4} + 4 x ^{2} + 3 )} \Rightarrow 1 + \frac{( - 4 x ^{2} - 3 )}{( x ^{4} + 4 x ^{2} + 3 )} \Rightarrow \frac{- 4 x ^{2} + 3}{( x ^{2} + 1 ) ( x ^{2} + 3 )} = \frac{4 x + B}{( x ^{2} + 1 )} + \frac{cx + D}{( x ^{2} + 3 )} So, \frac{x ^{4}}{( x ^{2} + 1 ) ( x ^{2} + 3 )} = 1 + \frac{Ax + B}{( x ^{2} + 1 )} + \frac{cx + D}{( x ^{2} + 3 )}

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