The domain of f(x) = cot-1 (x /√ x2-[x2]) x ∈ R is

Question

The domain of f(x) = cot-1 (x /√ x2-[x2]) x ∈ R is (A) R (B) R - 0  (C) R - ± , √ ∈ N  (D) none of these

Tardigrade Admin · Accepted Answer

Domain of cot-1 x is R and (x/√x2-[x2]) is defined if x2 ≠ [(x2)]. i,e., x2 is not integer (∵ x2 ≥ [x2]) Hence x2 ≠ non-negative integer i.e., 0 or +ve integer. Hence domain = R - (√ n: n ≥ 0,n ∈ Z)

Q. The domain of $f (x) = cot^{- 1} \frac{x}{x ^{2} - [ x ^{2} ]}$ $x \in$ R is

R

$R - {0}$

$R - {\pm, \in N}$

none of these

Domain of $cot^{- 1}$ x is R and $\frac{x}{x ^{2} - [ x ^{2} ]}$ is defined if $x^{2} \neq = [(x^{2})]$ . i,e., $x^{2}$ is not integer $(∵ x^{2} \geq [x^{2}])$
Hence $x^{2} \neq =$ non-negative integer i.e., 0 or $+ v e$ integer.
Hence domain = R $- {(n : n \geq 0, n \in Z)}$

Q. The domain of f(x)=cot−1x​2−[x2]x​ x∈ R is

R

R−{0}

R−{±,∈​N}

none of these

Domain of cot−1 x is R and x2−[x2]​x​ is defined if x2=[(x2)]. i,e., x2 is not integer (∵x2≥[x2]) Hence x2= non-negative integer i.e., 0 or +ve integer. Hence domain = R −{(n​:n≥0,n∈Z)}

Q. The domain of $f (x) = cot^{- 1} \frac{x}{x ^{2} - [ x ^{2} ]}$ $x \in$ R is

$R - {0}$

$R - {\pm, \in N}$

Domain of $cot^{- 1}$ x is R and $\frac{x}{x ^{2} - [ x ^{2} ]}$ is defined if $x^{2} \neq = [(x^{2})]$ . i,e., $x^{2}$ is not integer $(∵ x^{2} \geq [x^{2}])$
Hence $x^{2} \neq =$ non-negative integer i.e., 0 or $+ v e$ integer.
Hence domain = R $- {(n : n \geq 0, n \in Z)}$