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The area of the region bounded by the curves \[y=\left| x-1 \right|\] and \[y=3-\left| x \right|\] is     AIEEE  Solved  Paper-2003

A)                                         2 sq units        

B)                       3 sq units                            

C)       4 sq units        

D)                       6 sq units

Correct Answer: C

Solution :

\[y=\left| x-1 \right|=\left\{ \begin{matrix}    x-1, & x>0  \\    -x+1, & x0  \\ \end{matrix} \right.\]        Solving \[y=x-1\] and \[y=3-x\] \[\Rightarrow \]   \[x-1=3-x\] \[\Rightarrow \]   \[x=2\] and            \[y=3-2=1\]             Now, \[A{{B}^{2}}={{(2-1)}^{2}}+(1-{{0}^{2}})=1+1=2\] \[\Rightarrow \]   \[AB=\sqrt{2}\] and \[B{{C}^{2}}={{(0-2)}^{2}}+{{(3-1)}^{2}}=4+4=8\]                                     \[BC=2\sqrt{2}\] Area of rectangle ABCD = AB \[\times \] BC \[\Rightarrow \]   \[=\sqrt{2}\times 2\,\sqrt{2}=4\] sq units