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TS EAMCET 19-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

\frac{42-13x}{x^{2}+x-6} = \frac{A}{lx+m} + \frac{B}{px+q}

lm>0

p q < 0

\frac{A l p}{B m q} =

$\frac{27}{32}$
$\frac{27}{8}$
$\frac{8}{243}$
$\frac{243}{32}$

Solution: 👈: Video Solution

Let

\frac{42-13x}{x^2+x-6} = \frac{M}{x+3} + \frac{N}{x-2}

42-13x = M x - 2 M + N x + 3 N

\Rightarrow M+N = -13 \text{ and } -2M+3N = 42

-2(-13-N) + 3N = 42

26 + 2N + 3N = 42

5N = 16 \Rightarrow N = \frac{16}{5}

M = -13 - \frac{16}{5} = \frac{-65-16}{5} = -\frac{81}{5}

\frac{42-3x}{x^2+x-16} = \frac{-81}{5(x+3)} + \frac{16}{5(x-2)}

= \frac{-81}{5x+15} + \frac{16}{5x-10}

\text{ Here, } A=-81, B=16, l=p=5

m=15, q=-10

\frac{A l p}{B m q} = \frac{-81 \times 5 \times 5}{16 \times 15 \times (-10)} = \frac{27}{32}

\;

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