JEE MATHEMATICS MCQ: PARTIAL FRACTIONS

1 Easy

1 of 1

Application +4 | -1

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

, then

16A + 9B

is equal to

View answer and explanation

Correct option: C

Option C (6) is correct because after performing partial fraction decomposition and solving for A and B, substituting these values into 16A + 9B yields 6.

More Questions from PARTIAL FRACTIONS

Practice All Chapter MCQs →

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 If

\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}}

, then

A

is equal to If

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

, then

A - B

is equal to If

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

can be expanded in the ascending powers of

x

, then the coeficient of

{x^3}

is Let

a,b

and

c

be such that

\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}}

then

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

is equal to

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

is equal to