Question

f(x)= √log1/2(x2-2x+2), then domain of f(x) is

Answer 1

Function is $\bold{f(x)=\sqrt{log_{1/2}^{(x^2-2x+2)}}}$
conditions for defining function :
1. (x² - 2x + 2) > 0 [ for log to be defined ]
2. $\bold{log_{1/2}(x^2-2x+2)} \geq 0$ [ square root always positive ]

Now , solving inequality 1 ,
x² - 2x + 2 > 0
Here you see coefficient of x² = 1 > 0
But discriminant , D = b² - 4ac = (2)² - 4.2 = -4 < 0
∴ (x² - 2x + 2) > 0 for all real value of x
so, x ∈ R

Solving inequality 2 ,
$\bold{log_{1/2}(x^2-2x+2)} \geq 0$
⇒$\bold{-log_{2}(x^2-2x+2)} \geq 0$
⇒$\bold{log_{2}(x^2-2x+2)} \leq 0$
⇒ (x² - 2x + 2) ≤ 1
⇒ x² - 2x + 1 ≤ 0
⇒ (x - 1)² ≤ 0
∵ (x - 1)² ≠ negative
so, (x - 1)² = 0
⇒ x = 1

Now, take common value of x from both cases ,
x ∈ R and x ∈ {1 }
∴ x ∈ {1}

Hence, domain of f(x) ∈ { 1}