Q48 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
If $\frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2 i$, then $2 x+4 y=$
- Correct answerA5
- B-2
- C1
- D$-\mathbb{1}$
Solution
$\frac{(2-i) x+(1+i)}{(2+i)}+\frac{(1-2 i) y+(1-i)}{(1+2 i)}=1-2 i$
$\begin{aligned}
& \Rightarrow \frac{(2-i)^2 x+(1+i)(2-i)}{4-i^2}+\frac{(1-2 i)^2 y+(1-i)(1-2 i)}{1-4 i^2} \\
& \quad=1-2 i \\
& \Rightarrow \frac{\left(4+i^2-4 i\right) x+\left(2-i+2 i-i^2\right)}{5} \\
& \quad+\frac{\left(1+4 i^2-4 i\right) y+\left(1-2 i-i+2 i^2\right)}{5}=1-2 i \\
& \Rightarrow \frac{1}{5}[(3 x-3 y+2)+i(-48 x-4 y-2)]=1-2 i \\
& \Rightarrow \frac{1}{5}(3 x-3 y+2)-\frac{2}{5}(2 x+2 y+1) i=1-2 i
\end{aligned}$
On compare we get $\begin{aligned} & 3 x-3 y+2=5 \Rightarrow x-y=1 \qquad ....\mathrm{(i)} \\ & 2 x+2 y+1=5 \Rightarrow x+y=2 \qquad ....\mathrm{(ii)} \end{aligned}$
On solving (i) and (ii), we get $x=\frac{3}{2}$ and $y=\frac{1}{2}$ Now, $2 x+4 y=2\left(\frac{3}{2}\right)+4\left(\frac{1}{2}\right)=5$.
On compare we get $\begin{aligned} & 3 x-3 y+2=5 \Rightarrow x-y=1 \qquad ....\mathrm{(i)} \\ & 2 x+2 y+1=5 \Rightarrow x+y=2 \qquad ....\mathrm{(ii)} \end{aligned}$
On solving (i) and (ii), we get $x=\frac{3}{2}$ and $y=\frac{1}{2}$ Now, $2 x+4 y=2\left(\frac{3}{2}\right)+4\left(\frac{1}{2}\right)=5$.
Hence, the correct answer is option: 'A'.