AP EAPCET 08-Jul-2022 Shift 2 Paper

Given, f(x)=√

(2−|x|)

(3−|x|)

for f(x) to be defined,

2−|x|

3−|x|

≥0
Case I :
If x≥0, then

2−|x|

3−|x|

≥0 ⇒

x−2

x−3

≥0

Thus, x∊(−∞,2)υ(3,∞) (i)
Case II :
If x<0, then

2+x

3+x

≥0

Thus, x∊(−∞,−3)u(−2,∞) (ii)
Case III :
∵3−|x|≠0⇒|x|≠3
⇒x≠3,−3 (iii)

From eqs (i),(ii) and (iii) we get:
xε(−∞,−3)∪(−2,2)∪(3,∞)