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1)

Area  of the region {(x,y)}ε R2:yx+3,5y(x+9)15 } is equal to


A) 16

B) 43

C) 32

D) 53

Answer:

Option C

Explanation:

Here, {(x,y)}ε R2:yx+3,5y(x+9)15 }

\Rightarrow    y\geq\begin{cases}\sqrt{x+3},& when & x \geq -3\\\sqrt{-x-3}& when & x \leq-3\end{cases}

or           y^{2}\geq\begin{cases}x+3 & when& x \geq -3\\-3-x & when & x \leq -3\end{cases}

shown as

1422021305_c17.JPG

Also,  5y\leq (x+9)\leq 15

\Rightarrow (x+9)\geq 5y and x\leq 6

shown as 

1422021316_m1.JPG

\therefore      { (x,y)\epsilon R^{2}:y\geq\sqrt{\mid x+3 \mid}, 5y\leq(x+9)\leq15}

1422021842_c18.JPG

\therefore  Required area

                               = Area of trapezium ABCD - Area of ABE under parabola- Area of CDE under parabola

                         = \frac{1}{2}(1+2)(5)-\int_{-4}^{-3}\sqrt{-(x+3)}dx-\int_{-3}^{1}\sqrt{(x+3) } dx

                  =\frac{15}{2}-[\frac{(-3-x)^{\frac{3}{2}}}{-\frac{3}{2}}]_{-4}^{-3}-[\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}]^{1}_{-3}

    =\frac{15}{2}+\frac{2}{3}[0-1]-\frac{2}{3}[8-0]

=\frac{15}{2}-\frac{2}{3}-\frac{16}{3}=\frac{15}{2}-\frac{18}{3}=\frac{3}{2}