Q59 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

If $\frac{1}{(3 x+1)(x-2)}=\frac{A}{3 x+1}+\frac{B}{x-2}$ and $\frac{x+1}{(3 x+1)(x-2)}=\frac{C}{3 x+1}+\frac{D}{x-2}$, then

  • A
    $A+3 B=0, A: C=1: 3, B: D=2: 3$
  • B
    $A+3 B=0, A: C=3: 1, B: D=3: 2$
  • C
    $A-3 B=0, A: C=3: 2, B: D=1: 3$
  • Correct answer
    D
    $A+3 B=0, A: C=3: 2, B: D=1: 3$

Solution

$\begin{aligned} & \text { } \frac{1}{(3 x+1)(x-2)}=\frac{\mathrm{A}}{3 x+1}+\frac{\mathrm{B}}{x-2} \\ & \frac{1}{(3 x+1)(x-2)}=\left\{\frac{3}{3 x+1}-\frac{1}{x-2}\right\} \times\left(\frac{-1}{7}\right) \\ & =\frac{-\frac{3}{7}}{3 x+1}+\frac{\frac{4}{7}}{x-2} \Rightarrow \mathrm{~A}=\frac{-3}{7}, \mathrm{~B}=\frac{1}{7} \\ & \frac{x+1}{(3 x+1)(x-2)}=\frac{x-2+3}{(3 x+1)(x-2)} \\ & =\frac{1}{3 x+1}-\frac{3}{7}\left[\frac{3}{3 x+1}-\frac{1}{x-2}\right] \\ & =\frac{1}{3 x+1}-\frac{9}{7(3 x+1)}+\frac{\frac{3}{7}}{(x-2)}=\frac{\frac{-2}{7}}{3 x+1}+\frac{\frac{3}{7}}{x-2} \\ & \mathrm{~A}=\frac{-3}{7}, \mathrm{~B}=\frac{1}{7}, \mathrm{C}=\frac{-2}{7}, \mathrm{D}=\frac{3}{7} \\ & \mathrm{~A}+3 \mathrm{~B}=0, \mathrm{~A}: \mathrm{C}=3: 2, \mathrm{~B}: \mathrm{D}=1: 3 \end{aligned}$

Hence, the correct answer is option: 'D'.

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