Question

$\frac{\left(3x+4\right)}{{\left(x+1\right)}^2\left(x-1\right)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{{\left(x+1\right)}^2}$ Then $A=$

• A. $-\frac{1}{2}$
• B. $\frac{15}{4}$
• C. $\frac{7}{4}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is C

$\frac{7}{4}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{\left(3x+4\right)}{{\left(x+1\right)}^2\left(x-1\right)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{{\left(x+1\right)}^2}$

$$\begin{gathered} \Rightarrow \frac{\left(3x+4\right)}{{\left(x+1\right)}^2\left(x-1\right)}=\frac{A{\left(x+1\right)}^2+B\left(x^2-1\right)+C\left(x-1\right)}{{\left(x+1\right)}^2\left(x-1\right)} \\ \Rightarrow \frac{\left(3x+4\right)}{{\left(x+1\right)}^2\left(x-1\right)}=\frac{A\left(x^2+1+2x\right)+B\left(x^2-1\right)+C\left(x-1\right)}{{\left(x+1\right)}^2\left(x-1\right)} \\ \Rightarrow \frac{\left(3x+4\right)}{{\left(x+1\right)}^2\left(x-1\right)}=\frac{\left(A+B\right)x^2+\left(2A+C\right)+\left(A-B-C\right)}{{\left(x+1\right)}^2\left(x-1\right)} \end{gathered}$$

Step-2: Comparing coefficients.

$$\begin{gathered} A+B=0 \\ \Rightarrow A=-B..............\left(i\right) \\ 2A+C=3 \\ \Rightarrow C=3-2A..............\left(ii\right) \\ A-B-C=4..............\left(iii\right) \end{gathered}$$

Step-3: Solving equation (i), (ii) & (iii).

$$\begin{gathered} A-\left(-A\right)-\left(3-2A\right)=4 \\ \Rightarrow 4A=7 \\ \Rightarrow A=\frac{7}{4} \end{gathered}$$

Therefore, correct answer is option $\left(C\right)$