1. Given function

$f(x)=\sin^{-1}\big([2x^2-3]\big)+\log_2\left(\log_{1/2}(x^2-5x+5)\right)$

We need the domain of $f(x)$.

So both parts must be defined:

• $\sin^{-1}([2x^2-3])$ must exist
• $\log_2\left(\log_{1/2}(x^2-5x+5)\right)$ must exist

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2. Condition from the inverse sine term

For $\sin^{-1}(y)$ to be defined,

$-1 \le y \le 1$

Here,

$y=[2x^2-3]$

Since $[2x^2-3]$ is an integer, the only possible integer values between $-1$ and $1$ are:

$-1,0,1$

Thus,

$-1 \le 2x^2-3 < 2$

because $[t]\in\{-1,0,1\}$ means

$-1 \le t < 2$

So,

$-1 \le 2x^2-3 < 2$

Add $3$:

$2 \le 2x^2 < 5$

Divide by $2$:

$1 \le x^2 < \frac52$

Hence,

$x\in \left(-\sqrt{\frac52},-1\right]\cup\left[1,\sqrt{\frac52}\right)$

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3. Condition from the logarithmic term

We need

$\log_2\left(\log_{1/2}(x^2-5x+5)\right)$

to be defined.

For outer logarithm $\log_2(u)$, we need

$u>0$

So,

$\log_{1/2}(x^2-5x+5)>0$

Now base $\frac12$ is between $0$ and $1$, so $\log_{1/2}(t)>0$ implies

$0<t<1$

Therefore,

$0<x^2-5x+5<1$

This gives two inequalities:

(i) $x^2-5x+5>0$

Roots of $x^2-5x+5=0$ are

$x=\frac{5\pm\sqrt5}{2}$

Since the parabola opens upward,

$x^2-5x+5>0 \iff x<\frac{5-\sqrt5}{2} \quad \text{or} \quad x>\frac{5+\sqrt5}{2}$

(ii) $x^2-5x+5<1$

$x^2-5x+4<0$

$(x-1)(x-4)<0$

So,

$1<x<4$

Now intersect (i) and (ii):

$1<x<4$

with

$x<\frac{5-\sqrt5}{2} \quad \text{or} \quad x>\frac{5+\sqrt5}{2}$

Since

$\frac{5-\sqrt5}{2}\approx 1.382, \qquad \frac{5+\sqrt5}{2}\approx 3.618$

we get

$x\in\left(1,\frac{5-\sqrt5}{2}\right)\cup\left(\frac{5+\sqrt5}{2},4\right)$

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4. Intersect both domain conditions

From inverse sine:

$x\in \left(-\sqrt{\frac52},-1\right]\cup\left[1,\sqrt{\frac52}\right)$

From logarithm:

$x\in\left(1,\frac{5-\sqrt5}{2}\right)\cup\left(\frac{5+\sqrt5}{2},4\right)$

Now,

$\sqrt{\frac52}\approx 1.581$

So only the interval

$\left(1,\frac{5-\sqrt5}{2}\right)$

lies inside $\left[1,\sqrt{\frac52}\right)$.

The interval $\left(\frac{5+\sqrt5}{2},4\right)$ does not intersect because

$\frac{5+\sqrt5}{2}>\sqrt{\frac52}$

Hence the final domain is

$\boxed{\left(1,\frac{5-\sqrt5}{2}\right)}$

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5. Match with options

This is Option C.

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6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.