question_answer

The area bounded by  \[y-1=\left| x \right|,\,\,y=0\,\,and\,\,\left| x \right|=\frac{1}{2}\] will be:

A) \[\frac{3}{4}\]                        

B)   \[\frac{3}{2}\]

C) \[\frac{5}{4}\]                        

D)   None of these

Correct Answer: C

Solution :

The given lines are, \[\operatorname{y}-1=x,\,\,x\ge 0;\,\,y\,-1\,\,=\,\,-x,\,\,x<0\] \[\operatorname{y}=\,0;\,\,\,x=-\frac{1}{2},\,\,x<0; \,\,x=\frac{1}{2}, \,x\ge 0\] so that the area bounded is as shown in the figure. Required area \[=\,\,2\int_{0}^{1/2}{(1+x)dx}=\,\,2\left( x+\frac{{{x}^{2}}}{2} \right)_{0}^{1/2}\,\,=\,\,2\left( \frac{1}{2}+\frac{1}{8} \right)\,\,=\,\,\frac{5}{4}\]