The domain of the function $f(x)=\sin ^{-1}\left[2 x^{2}-3\right]+\log _{2}\left(\log _{\frac{1}{2}}\left(x^{2}-5 x+5\right)\right)$, where $[t]$ is the greatest integer function, is :

(C) $\left(1, \frac{5-\sqrt{5}}{2}\right)$

(A) $\left(-\sqrt{\frac{5}{2}}, \frac{5-\sqrt{5}}{2}\right)$

(B) $\left(\frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)$

(D) $\left[1, \frac{5+\sqrt{5}}{2}\right)$

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Sets Relations and Functions

The domain of the function

f (x) = sin^{- 1} [2 x^{2} - 3] + lo g_{2} (lo g_{\frac{1}{2}} (x^{2} - 5 x + 5))

,
where

[t]

is the greatest integer function, is :

Answer: (C)

For

f (x)

to be defined, we need three conditions:

Condition 1: For

sin^{- 1} [2 x^{2} - 3]

to exist:

- 1 \leq [2 x^{2} - 3] \leq 1

. Since

[2 x^{2} - 3]

is an integer, we need

- 1 \leq 2 x^{2} - 3 < 2

, giving

1 < x^{2} < \frac{5}{2}

, so

x \in (- \frac{5}{2}, - 1) \cup (1, \frac{5}{2})

.

Condition 2: For the logarithm to exist:

x^{2} - 5 x + 5 > 0

. The roots are

\frac{5 \pm 5}{2}

, so

x \in (- \infty, \frac{5 - 5}{2}) \cup (\frac{5 + 5}{2}, \infty)

.

Condition 3: For

lo g_{2} (lo g_{1/2} (x^{2} - 5 x + 5))

to exist:

lo g_{1/2} (x^{2} - 5 x + 5) > 0

, which means

x^{2} - 5 x + 5 < 1

, giving

x^{2} - 5 x + 4 < 0

, so

x \in (1, 4)

.

Intersecting all three conditions:

(1, \frac{5 - 5}{2})

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option