The domain of the function f defined by f(x)=(1/√x-|x|) is

Question

The domain of the function f defined by f(x)=(1/√x-|x|) is (A) R (B) R+ (C) R- (D)  φ

Tardigrade Admin · Accepted Answer

Given that f(x)=(1/√x-|x|) where x-|x| = begincases x-x=0, textif x ge0 [2ex] x-(-x)=2x, textif x<0 endcases Thus (1/√x-|x|) is not defined for any x ∈ R. Hence f is not defined for any x ∈ R i.e., domain of f= φ .

Q. The domain of the function $f$ defined by $f (x) = \frac{1}{x - ∣ x ∣}$ is

$R$

$R^{+}$

$R^{-}$

${ϕ}$

Given that $f (x) = \frac{1}{x - ∣ x ∣}$
where $x - ∣ x ∣ = ⎩ ⎨ ⎧ x - x = 0, x - (- x) = 2 x, if x \geq 0 if x < 0$
Thus $\frac{1}{x - ∣ x ∣}$ is not defined for any $x \in R$ .
Hence $f$ is not defined for any $x \in R$
i.e., domain of $f = {ϕ}$ .

Q. The domain of the function f defined by f(x)=x−∣x∣​1​ is

R

R+

R−

{ϕ}

Given that f(x)=x−∣x∣​1​ where x−∣x∣=⎩⎨⎧​x−x=0,x−(−x)=2x,​if x≥0if x<0​ Thus x−∣x∣​1​ is not defined for any x∈R. Hence f is not defined for any x∈R i.e., domain of f={ϕ}.

Q. The domain of the function $f$ defined by $f (x) = \frac{1}{x - ∣ x ∣}$ is

$R$

$R^{+}$

$R^{-}$

${ϕ}$

Given that $f (x) = \frac{1}{x - ∣ x ∣}$
where $x - ∣ x ∣ = ⎩ ⎨ ⎧ x - x = 0, x - (- x) = 2 x, if x \geq 0 if x < 0$
Thus $\frac{1}{x - ∣ x ∣}$ is not defined for any $x \in R$ .
Hence $f$ is not defined for any $x \in R$
i.e., domain of $f = {ϕ}$ .