Question

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 1

The correct option is C $-\frac{4}{3},\frac{11}{3}$

$\frac{5x+6}{(2+x)(1-x)}=\frac{a}{2+x}+\frac{b}{1-x}⟹\frac{5x+6}{(2+x)(1-x)}=\frac{a-ax+2b+bx}{(2+x)(1-x)}⟹\frac{5x+6}{(2+x)(1-x)}=\frac{x(b-a)+(a+2b)}{(2+x)(1-x)}$

Equating the coefficient of x and the constant term of both the sides,

$b-a=5$ .....(i)

$a+2b=6$ .......(ii)

Adding (i) & (ii), we get

$3b=11$

$⟹b=\frac{11}{3}$ .

Putting $b=\frac{11}{3}$ in (i),

$-a=5-b=5-\frac{11}{3}=\frac{4}{3}.\therefore a=-\frac{4}{3}.\therefore (a,b)=(-\frac{4}{3},\frac{11}{3})$