AP EAPCET 24-Aug-2021 Shift 1 Paper

f(x)=√

1−|x|

2−|x|

For domain

1−|x|

2−|x|

≥0 ...(i)
and 2−|x|≠0
|x|≠2
x=±2
Case I When, x≥0
So, Eq. (i) becomes

1−x

2−x

≥0‌‌‌‌ {∵|x|=[

x	x≥0
−x	x<0

}
Now, critical points are
x=1,2
Using wavy curve method But x≥0
Solution is x∈[0,1]∪(2,∞).
Case II When, x<0
So, Eq. (i) becomes

1+x

2+x

≥0
Critical points are x=−1,−2 Using wavy curve method But x<0
Solution is x∈(−∞,−2)∪[−1,0)
∴ Required solution is union of case I and case Il. ∴ Required solution is
x∈(−∞,−2)∪[−1,1]∪(2,∞)