JEE MATHEMATICS MCQ: PARTIAL FRACTIONS

1 Easy

1 of 1

Application +4 | -1

If a real valued function

f

of a real variable

x

is such that

\frac{1}{{\left( {1 + x} \right)\left( {1 + {x^2}} \right)}} = \frac{A}{{1 + x}} + \frac{{f\left( x \right)}}{{1 + {x^2}}}

then

f\left( x \right)

is equal to

View answer and explanation

Correct option: A

\frac{{1 - x}}{2}

Option A (

\frac{{1 - x}}{2}

) is correct because by performing partial fraction decomposition, we can find the function

f(x)

that satisfies the given equation.

More Questions from PARTIAL FRACTIONS

Practice All Chapter MCQs →

\frac{{{x^3}}}{{\left( {2x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)}} = A + \frac{B}{{2x - 1}} + \frac{C}{{x + 2}} + \frac{D}{{x - 3}}

A

\frac{{{x^2} + x + 1}}{{{x^2} + 2x + 1}} = A + \frac{B}{{x + 1}} + \frac{C}{{{{\left( {x + 1} \right)}^2}}}

A - B

\frac{{x - 4}}{{{x^2} - 5x + 6}}

x

{x^3}

a,b

c

\frac{1}{{\left( {1 - x} \right)\left( {1 - 2x} \right)\left( {1 - 3x} \right)}} = \frac{a}{{1 - x}} + \frac{b}{{1 - 2x}} + \frac{c}{{1 - 3x}}

\frac{a}{1} + \frac{b}{3} + \frac{c}{5}

\frac{{3{x^2} + 1}}{{{x^2} - 6x + 8}}

\frac{3 x^{3}-8 x^{2}+10}{(x-1)^{4}}