The domain of the function $\operatorname{cosec}^{-1}\left(\frac{1+x}{x}\right)$ is:

(B) $\left[-\frac{1}{2}, \infty\right)-\{0\}$

(A) $\left(-1,-\frac{1}{2}\right] \cup(0, \infty)$

(C) $\left[-\frac{1}{2}, 0\right) \cup[1, \infty)$

(D) $\left(-\frac{1}{2}, \infty\right)-\{0\}$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Inverse Trigonometric Functions

The domain of the function

cosec^{- 1} (\frac{1 + x}{x})

is:

Answer: (B)

The inverse cosecant is defined only when the modulus of its argument is at least

1

, and the argument itself requires

x \neq = 0

. So

\frac{1 + x}{x} \geq 1 ⟺ ∣1 + x ∣ \geq ∣ x ∣

Both sides being non-negative, squaring is reversible:

1 + 2 x + x^{2} \geq x^{2}

, that is

x \geq - \frac{1}{2}

. At

x = - \frac{1}{2}

the argument is

- 1

, which is admissible. Combining with

x \neq = 0

, the domain is

[- \frac{1}{2}, \infty) - {0}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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