<p>The area enclosed by the curves \(y=3-x\), \(y=x+1\) and \(y=0\) in the first quadrant is: [MAU052]</p>
Step-by-Step Solution
Key Concept: Lines y=3-x and y=x+1 meet at (1,2). With y=0: triangle vertices (0,0)? Find the triangular region.
<div class='solution'>
<p>Intersections: $3-x=x+1\Rightarrow x=1, y=2$. With y=0: $3-x=0\Rightarrow x=3$; $x+1=0\Rightarrow x=-1$ (outside 1st quadrant). $y=0$ meets x-axis.</p>
<p>In first quadrant, region bounded by the two lines and y=0: vertices at $(3,0)$, $(0,1)$ (from $y=x+1$ at $x=0$... $y=1$), and $(1,2)$.</p>
<p>Area of this triangular region $=2$. ✓ (B)</p>
</div>
Correct Answer: ['B']