Inverse Trigonometric Functions

Given function,
f(x)=√loge(‌

1

x2−4x+4

)+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]