AP EAMCET · Maths · Functions
For \(x\) and \(y\) satisfying \(|x|+|y|=|x-3|+|y-2|\), which of the following is correct?
- A \(x=\frac{1}{2}\) for \(0 \leq x < 3\) and \(1 \leq y < 2\)
- B \(x+y=\frac{5}{2}\) for \(x \geq 3\) and \(y \geq 2\)
- C \(x=\frac{1}{2}\) for \(x \geq 2\) and \(0 \leq y < 3\)
- D \(x+y=\frac{5}{2}\) for \(0 \leq x < 3,0 \leq y < 2\)
Answer & Solution
Correct Answer
(D) \(x+y=\frac{5}{2}\) for \(0 \leq x < 3,0 \leq y < 2\)
Step-by-step Solution
Detailed explanation
\(\because|x|+|y|=|x-3|+|y-2|\) Case I \(0 \leq x < 3\) and \(0 \leq y < 2\) \(|x|+|y|=|x-3|+|y-2|\) \(\Rightarrow \quad x+y=-(x-3)-(y-2)\) \(\begin{aligned} & \Rightarrow \quad x+y=-x+3-y+2 \\ & \Rightarrow \quad 2 x+2 y=5 \Rightarrow x+y=5 / 2\end{aligned}\) Case II…
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