The domain of the function cos -1((2 sin -1((1/4 x2-1))/π)) is :

Question

The domain of the function cos -1((2 sin -1((1/4 x2-1))/π)) is : (A) R- -(1/2), (1/2)  (B) (-∞,-1] ∪[1, ∞) ∪ 0  (C) (-∞, (-1/2)) ∪((1/2), ∞) ∪ 0  (D) (-∞, (-1/√2)] ∪[(1/√2), ∞) ∪ 0

Tardigrade Admin · Accepted Answer

-1 ≤ (2 sin -1((1/4 x2-1))/π) ≤ 1 -π / 2 ≤ sin -1 (1/4 x2-1) ≤ π / 2 Always -1 ≤ (1/4 x2-1) ≤ 1 x ∈(∞, (1/√2)) ∪[(1/√2), ∞)

Q. The domain of the function $cos^{- 1} (\frac{2 s i n ^{- 1} ( \frac{1}{4 x ^{2} - 1} )}{π})$ is :

$R - {- \frac{1}{2}, \frac{1}{2}}$

$(- \infty, - 1] \cup [1, \infty) \cup {0}$

$(- \infty, \frac{- 1}{2}) \cup (\frac{1}{2}, \infty) \cup {0}$

$(- \infty, \frac{- 1}{2}] \cup [\frac{1}{2}, \infty) \cup {0}$

$- 1 \leq \frac{2 s i n ^{- 1} ( \frac{1}{4 x ^{2} - 1} )}{π} \leq 1$
$- π /2 \leq sin^{- 1} \frac{1}{4 x ^{2} - 1} \leq π /2$
Always $- 1 \leq \frac{1}{4 x ^{2} - 1} \leq 1$
$x \in (\infty, \frac{1}{2}) \cup [\frac{1}{2}, \infty)$

Q. The domain of the function cos−1(π2sin−1(4x2−11​)​) is :

R−{−21​,21​}

(−∞,−1]∪[1,∞)∪{0}

(−∞,2−1​)∪(21​,∞)∪{0}

(−∞,2​−1​]∪[2​1​,∞)∪{0}