Let R be the region represented by |x−y|+|x+y|≤8.Then

a), b) Required area =2∫1/24(4−2/x)dx=2(4x−2logx)1/24=4(7−3logx)=4(7−log38)c) Area of region R=Area of square =8×8=64d) Area of region bounded by xy≤2 and R is Area of square – area bounded by xy≤2 and R 64−4(7−loge8)=12(3+loge2)

Let R be the region represented by |x−y|+|x+y|≤8.Then

a), b) Required area =2∫1/24(4−2/x)dx=2(4x−2logx)1/24=4(7−3logx)=4(7−log38)c) Area of region R=Area of square =8×8=64d) Area of region bounded by xy≤2 and R is Area of square – area bounded by xy≤2 and R 64−4(7−loge8)=12(3+loge2)

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Let R be the region represented by $| x - y | + | x + y | \leq 8.$ Then

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Area bounded by $x y \geq 2$ and R is $4 (7 - \log_{e} 8)$

Area bounded by $x y \geq - 2$ and R is $4 (7 - \log_{e} 8)$

Area of the region R is 64

Area bounded by $x y \leq 2$ and R is $12 (3 + \log_{e} 2)$

answer is A, B, C, D.

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Detailed Solution

a), b) Required area $= 2 \int_{1 / 2}^{4} (4 - 2 / x) d x$

$= 2 {(4 x - 2 \log x)}_{1 / 2}^{4}$

$= 4 (7 - 3 \log x) = 4 (7 - \log_{3} 8)$

c) Area of region R=Area of square $= 8 \times 8 = 64$

d) Area of region bounded by $x y \leq 2$ and R is Area of square – area bounded by $x y \leq 2$ and R $64 - 4 (7 - \log_{e} 8) = 12 (3 + \log_{e} 2)$