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Question: If \(\frac{3x + 4}{(x + 1)^{2}(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^{2...

If 3x+4(x+1)2(x−1)=A(x−1)+B(x+1)+C(x+1)2\frac{3x + 4}{(x + 1)^{2}(x - 1)} = \frac{A}{(x - 1)} + \frac{B}{(x + 1)} + \frac{C}{(x + 1)^{2}} and 3x+4=A(x+1)23x + 4 = A(x + 1)^{2} then.

A

A(2)2A(2)^{2}

B

A=74A = \frac{7}{4}

C

3x−1(1−x+x2)(2+x)=Ax+Bx2−x+1+Cx+2\frac{3x - 1}{(1 - x + x^{2})(2 + x)} = \frac{Ax + B}{x^{2} - x + 1} + \frac{C}{x + 2}

D

None of these

Answer

A(2)2A(2)^{2}

Explanation

Solution

(21−18)−(20−17)(\sqrt{21} - \sqrt{18}) - (\sqrt{20} - \sqrt{17})

(21−18)(21+18)21+18−20−1720+17\frac{(\sqrt{21} - \sqrt{18})(\sqrt{21} + \sqrt{18})}{\sqrt{21} + \sqrt{18}} - \frac{20 - 17}{\sqrt{20} + \sqrt{17}}

As 3[121+18−120+17]3\left\lbrack \frac{1}{\sqrt{21} + \sqrt{18}} - \frac{1}{\sqrt{20} + \sqrt{17}} \right\rbrack i.e., 3[20+17−21−18](21+18)(20+17)\frac{3\lbrack\sqrt{20} + \sqrt{17} - \sqrt{21} - \sqrt{18}\rbrack}{(\sqrt{21} + \sqrt{18})(\sqrt{20} + \sqrt{17})}. 3[(20−21)+(17−18)](21+18)(20+17)\frac{3\lbrack(\sqrt{20} - \sqrt{21}) + (\sqrt{17} - \sqrt{18)}\rbrack}{(\sqrt{21} + \sqrt{18})(\sqrt{20} + \sqrt{17})} −3[(21−20)+(18−17)(21+18)(20+17)<0\frac{- 3\lbrack(\sqrt{21} - \sqrt{20}) + (\sqrt{18} - \sqrt{17})}{(\sqrt{21} + \sqrt{18})(\sqrt{20} + \sqrt{17})} < 0

And ∴\therefore i.e., a<ba < b

x+10+x−2=6\sqrt{x + 10} + \sqrt{x - 2} = 6 ⇒\Rightarrow; x+10=6−x−2\sqrt{x + 10} = 6 - \sqrt{x - 2} ⇒\Rightarrow.