2
-2
1
-1
Correct answer is A
\(\frac{3x + 4}{(x - 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x - 2}\)
\(\frac{3x + 4}{(x - 2)(x + 3)}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}\)
=\(3x+4=P(x-2)+Q(x+3)\)
=\(3x+4=Px-2P+Qx+3Q\)
=\(3x+4=Px+Qx-2P+3Q\)
=\(3x+4=(P+Q)x-2P+3Q\)
Equating x and the constants
P+Q=3------(i)
2P+3Q=4-------(ii)
From equation (i)
P=3-Q------(iii)
Substitute (3-Q) for P inequation (ii)
=-2(3-Q)+3Q=4
=-6+2Q+3Q=4
=-6+5Q=4
=5Q=4+6
=5Q=10
∴Q=\(\frac{10}{5}=2\)
If \(f(x) = mx^{2} - 6x - 3\) and \(f'(1) = 12\), find the value of the constant m....
If \(Px^{2} + (P+1)x + P = 0\) has equal roots, find the values of P....
If \(B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\), find \(B^{-...
Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y....
Given that \(\log_{3}(x - y) = 1\) and \(\log_{3}(2x + y) = 2\), find the value of x...