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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

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

\frac{Ax+B}{(x+3)^2} + \frac{C}{x-2}

then 25 (B + 8C - A) =

Solution:

Consider the expression,

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

\frac{Ax+B}{(x+3)^2} + \frac{C}{x-2}

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

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

8=\frac{(A x+B)(x-2)+C(x+3)^{2}}{(x+3)^{2}(x-2)}

Putting x = 2, we get

C=\frac{8}{25}

Putting x = 0, we get

8=B(-2)+C(9)

=-2B+\frac{8}{25}(9)

B=\frac{-64}{25}

Putting x =1, we get

8=(A+B)(-1)+C(4)^{2}

8=-A-B+16C

8=-A+\frac{64}{25}+\frac{128}{25}

A=\frac{-8}{25}

Now,

25(B+8C-A)=25\left(\frac{-64}{25}+\frac{64}{25}+\frac{8}{25}\right)

= 8

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