If frac{42-13x}{x^2+x-6}=frac{A}{lx+m}+frac{B | vedclass

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(A) Given, $\frac{42-13x}{x^2+x-6} = \frac{A}{lx+m} + \frac{B}{px+q}$.
Factorizing the denominator: $x^2+x-6 = (x+3)(x-2)$.
So, $\frac{42-13x}{(x+3)

Vedclass · Accepted Answer

$\frac{27}{32}$

Vedclass · Answer

(A) Given, $\frac{42-13x}{x^2+x-6} = \frac{A}{lx+m} + \frac{B}{px+q}$.
Factorizing the denominator: $x^2+x-6 = (x+3)(x-2)$.
So, $\frac{42-13x}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$.
Using partial fractions: $42-13x = A(x-2) + B(x+3)$.
For $x=2$: $42-26 = 5B \implies 16 = 5B \implies B = \frac{16}{5}$.
For $x=-3$: $42+39 = -5A \implies 81 = -5A \implies A = -\frac{81}{5}$.
Comparing with $\frac{A}{lx+m} + \frac{B}{px+q}$, we have $l=1, m=3, p=1, q=-2$ (satisfying $lm=3 > 0$ and $pq=-2 < 0$).
Then, $\frac{Alp}{Bmq} = \frac{(-\frac{81}{5}) 	imes 1 	imes 1}{(\frac{16}{5}) 	imes 3 	imes (-2)} = \frac{-81/5}{-96/5} = \frac{81}{96} = \frac{27}{32}$.

If $\frac{42-13x}{x^2+x-6}=\frac{A}{lx+m}+\frac{B}{px+q}$ where $lm > 0$ and $pq < 0$, then $\frac{Alp}{Bmq} =$

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