If \(\frac{5X+6}{(2+X)(1-X)}\) = \(\frac{a}{2+X}+\frac{b}{1-X}\),

1.	If \(\frac{5X+6}{(2+X)(1-X)}\) = \(\frac{a}{2+X}+\frac{b}{1-X}\), then the values of a and b respectively are :(a) \(-\frac{5}{3},\frac{6}{5}\)(b) \(\frac{5}{3},-\frac{6}{5}\)(c) \(-\frac{4}{3},\frac{11}{3}\)(d) \(\frac{4}{3},-\frac{11}{3}\)
Answer» (c) \(-\frac{4}{3},\frac{11}{3}\) \(\frac{5X+6}{(2+X)(1-X)}\) = \(\frac{a}{(2+X)}\) + \(\frac{b}{(1-X)}\) \(\Rightarrow\) \(\frac{a(1-X)+b(2+X)}{(2+X)(1-X)}\) = \(\frac{5X+6}{(2+X)(1-X)}\) \(\Rightarrow\)\(\frac{a-aX+2b+bX}{(2+X)(1-X)}\) = \(\frac{5X+6}{(2+X)(1-X)}\) \(\Rightarrow\) \(\frac{(a+2b)-X(a-b)}{(2+X)(1-X)}\) = \(\frac{5X+6}{(2+X)(1-X)}\) Equating the coefficient of x and constant term on both the sides of the equation, we get a + 2b = 6 ......…(i) a – b = –5 ...........(ii) Subtracting eqn (ii) from eqn (i), we get 3b = 11 \(\Rightarrow\) b = \(\frac{11}{3}\) Putting the value of b in (ii), we get a - \(\frac{11}{3}\) = -5 \(\Rightarrow\)a = -5 + \(\frac{11}{3}\) = \(\frac{-15+11}{3}\) = \(-\frac{4}{3}\).

If \(\frac{5X+6}{(2+X)(1-X)}\) = \(\frac{a}{2+X}+\frac{b}{1-X}\), then the values of a and b respectively are :(a) \(-\frac{5}{3},\frac{6}{5}\)(b) \(\frac{5}{3},-\frac{6}{5}\)(c) \(-\frac{4}{3},\frac{11}{3}\)(d) \(\frac{4}{3},-\frac{11}{3}\)

Answer»

(c) \(-\frac{4}{3},\frac{11}{3}\)

\(\frac{5X+6}{(2+X)(1-X)}\) = \(\frac{a}{(2+X)}\) + \(\frac{b}{(1-X)}\)

\(\Rightarrow\) \(\frac{a(1-X)+b(2+X)}{(2+X)(1-X)}\) = \(\frac{5X+6}{(2+X)(1-X)}\)

\(\Rightarrow\)\(\frac{a-aX+2b+bX}{(2+X)(1-X)}\) = \(\frac{5X+6}{(2+X)(1-X)}\)

\(\Rightarrow\) \(\frac{(a+2b)-X(a-b)}{(2+X)(1-X)}\) = \(\frac{5X+6}{(2+X)(1-X)}\)

Equating the coefficient of x and constant term on both the sides of the equation, we get

a + 2b = 6 ......…(i)

a – b = –5 ...........(ii)

Subtracting eqn (ii) from eqn (i), we get

3b = 11

\(\Rightarrow\) b = \(\frac{11}{3}\)

Putting the value of b in (ii), we get

a - \(\frac{11}{3}\) = -5

\(\Rightarrow\)a = -5 + \(\frac{11}{3}\) = \(\frac{-15+11}{3}\) = \(-\frac{4}{3}\).