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The area of the bounded region enclosed by the curve $\mathrm{y=3-\left|x-\frac{1}{2}\right|-|x+1| \,\,and \: the \: x-axis}$ is:
Option: 1
$\frac{9}{4}$

Option: 2
$\frac{45}{16}$

Option: 3
$\frac{27}{8}$

Option: 4
$\frac{63}{16}$

Answers (1)

best_answer

$y=3-\left|x-\frac{1}{2}\right|-|x+1|$

$\mathrm{For \: x<-1}$

$\mathrm{y =3-\left(\frac{1}{2}-x\right)-(-1-x) }\\$

$\mathrm{=3+x-\frac{1}{2}+x+1} \\$

$\mathrm{=2 x+\frac{7}{2}}$

$\mathrm{For \: -1 \leq x<\frac{1}{2} }\\$

$\mathrm{y=3+x-\frac{1}{2}-x-1=\frac{3}{2}} \\$

$\mathrm{For \: x \geqslant \frac{1}{2}}\\$

$\mathrm{y=3-x+\frac{1}{2}-x-1=-2 x+\frac{5}{2}}$

$\mathrm{\text { Area } =\frac{1}{2}\left(\frac{3}{2} \times \frac{3}{4}\right)+\frac{3}{2} \times \frac{3}{2}+\frac{1}{2}\left(\frac{3}{2} \times \frac{3}{4}\right)} \\$

$\mathrm{=\frac{9}{16}+\frac{9}{4}+\frac{9}{16}} \\$

$\mathrm{=\frac{9}{8}+\frac{9}{4} }\\$

$\mathrm{=\frac{27}{8} }$

Hence the correct answer is option 3.

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