)+sin‌−1(x2−2) Since, log is defined for positive values only. So, ‌
1
x2−4x+4
>0 Also since, x2−4x+4=(x−2)2≥0 this holds for x≠2 ⇒‌‌ln(‌
1
(x−2)2
)≥0 Now, we know that ln(t) is strictly increasing for t>0 and ln(1)=0, the inequality ln‌
1
(x−2)2
≥0 is equivalent to ‌‌
1
(x−2)2
≥1⇔(x−2)2≤1 ‌⇒‌‌|x−2|≤1 i.e., ‌‌1≤x≤3 But log is undefined for x=2 ∴x∈[1,3]−{2} For sin‌−1(x2−2) to be real, x2−2∈[−1,1] ‌⇒‌‌−1≤x2−2≤1 ‌⇒‌‌1≤x2≤3⇒1≤|x|≤√3 ‌⇒‌‌x∈[−√3,−1]∪[1,√3]⇒x∈[1,√3]
