Question - Vedclass

English

Gujarati

Hindi

Basic of Logarithms

easy

If ${{3x + 4} \over {{x^2} - 3x + 2}} = {A \over {x - 2}} - {B \over {x - 1}}$,then $(A,\,B) = $

A

$(7, 10)$

B

$(10, 7)$

C

$(10, -7)$

D

$(-10, 7)$

Solution

(b) $3x + 4 = A(x – 1) – B(x – 2)$

$ \Rightarrow $ $3=A-B,$ $4 = – A + 2B$

$ \Rightarrow $ $A = 10,\,\,\,B = 7$

$\therefore (A,\,B) = (10,\,7)$.

Std 11

Mathematics

Share

Similar Questions

If ${{a{x^2} + bx + c} \over {(x – 1)\,(x + 2)\,(2x + 3)}}$=${3 \over {x – 1}} + {2 \over {x + 2}} – {5 \over {2x + 3}}$, then

easy

${1 \over {x({x^2} + 1)}} = {A \over x} + {{Bx + C} \over {({x^2} + 1)}}$, then $(A,\,B,\,C) = $

easy

(IIT-1995)

The coefficient of ${x^5}$ in the expansion of ${{{x^2} + 1} \over {({x^2} + 4)(x – 2)}}$ is

hard

If ${x \over {(x – 1)\,{{({x^2} + 1)}^2}}} = {1 \over 4}\left[ {{1 \over {(x – 1)}} – {{x + 1} \over {{x^2} + 1}}} \right] + y$ then $y =$

medium

If ${{2x} \over {{x^3} – 1}} = {A \over {x – 1}} + {{Bx + C} \over {{x^2} + x + 1}}$, then

medium