Domain of Function Operations

Given,

$$f(x)=\sin^{-1}[2x^2-3]+\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$$

For the function to exist, both parts must be defined.

First consider

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

Since $$\sin^{-1}t$$ is defined for $$-1\le t\le1,$$

we need $$-1\le[2x^2-3]\le1$$

As the GIF gives only integers,

$$[2x^2-3]\in\{-1,0,1\}$$

Hence,

$$-1\le2x^2-3<2$$

$$2\le2x^2<5$$

$$1\le x^2<\frac52$$

Therefore,

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

Now consider

$$\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$$

For the outer logarithm,

$$\log_{\frac12}(x^2-5x+5)>0$$

Since the base

$$\frac12<1,$$

$$\log_{\frac12}t>0\iff0<t<1$$

Thus,

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

Now solve separately.

From $$x^2-5x+5>0,$$

roots are $$x=\frac{5\pm\sqrt5}{2}$$

Hence,

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

Also,

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

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

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

$$1<x<4$$

Combining both,

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

Now intersect with

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

Since

$$\sqrt{\frac52}<\frac{5+\sqrt5}{2},$$

only the first interval survives.

Therefore, the domain is

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

The function is:

$$f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}$$

We need to evaluate the constraints for both the numerator and the denominator independently, and then find their intersection.

1. Constraints for the Numerator

For the logarithmic function $$\log_{(x+1)}(x-2)$$ to be defined, the following conditions must be met:

• The argument must be strictly positive:

  $$x - 2 > 0 \implies x > 2$$

• The base must be strictly positive:

  $$x + 1 > 0 \implies x > -1$$

• The base cannot equal 1:

  $$x + 1 \neq 1 \implies x \neq 0$$

Intersecting these three conditions for the numerator, we get $$x > 2$$.

2. Constraints for the Denominator

The denominator must not equal zero, and any functions within it must also be defined.

• The argument of the logarithm in the exponent must be strictly positive:

  $$x > 0$$

• The denominator cannot be zero:

  $$e^{2\log_e x} - (2x+3) \neq 0$$

Using the logarithm power rule, $$2\log_e x = \log_e(x^2)$$. For $$x > 0$$, the expression simplifies using the property $$e^{\log_e(a)} = a$$:

$$e^{\log_e(x^2)} = x^2$$

Substitute this back into the denominator inequality:

$$x^2 - (2x + 3) \neq 0$$
$$x^2 - 2x - 3 \neq 0$$

Factoring the quadratic equation gives:

$$(x - 3)(x + 1) \neq 0$$

This means $$x \neq 3$$ and $$x \neq -1$$.

3. Final Domain

To find the domain of $$f(x)$$, we take the intersection of the valid sets for both the numerator and the denominator:

• From the numerator: $$x > 2$$
• From the denominator: $$x > 0$$, $$x \neq 3$$, and $$x \neq -1$$

Combining these constraints, $$x$$ must be strictly greater than $$2$$, but it cannot equal $$3$$.

Therefore, the domain of $$f(x)$$ is:

$$x \in (2, 3) \cup (3, \infty)$$ or $$(2, \infty) \setminus \{3\}$$.

The domain of $$f(x) = 4^{-x^{2}} + \cos^{-1}\!\left(\frac{x}{2}-1\right) + \log(\cos x)$$ is the set of all real numbers that satisfy the individual domain conditions of every term appearing in the expression.

Since the question itself restricts us to $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$, we will enforce this outer restriction at the end. Let us examine each term one by one.

1. The exponential term $$4^{-x^{2}}$$
The base 4 is positive, so $$4^{(\text{anything})}$$ is defined for every real $$x$$. Hence this term imposes no extra restriction.

2. The inverse-cosine term $$\cos^{-1}\!\left(\frac{x}{2}-1\right)$$
For the principal value of $$\cos^{-1}(y)$$ to be defined, its argument must satisfy $$-1 \le y \le 1$$. Therefore
$$-1 \le \frac{x}{2}-1 \le 1.$$

Solving these two inequalities:

Lower bound:
$$\frac{x}{2}-1 \ge -1 \quad\Longrightarrow\quad \frac{x}{2} \ge 0 \quad\Longrightarrow\quad x \ge 0.$$

Upper bound:
$$\frac{x}{2}-1 \le 1 \quad\Longrightarrow\quad \frac{x}{2} \le 2 \quad\Longrightarrow\quad x \le 4.$$

Hence from the inverse-cosine term we obtain the interval
$$0 \le x \le 4.$$

3. The logarithmic term $$\log(\cos x)$$
The natural logarithm is defined only for positive arguments, so we need
$$\cos x \gt 0.$$

Inside the given outer interval $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$, the cosine function is positive for every point except the endpoints, where $$\cos\!\left(\pm\frac{\pi}{2}\right)=0$$. Therefore
$$\cos x \gt 0 \quad\Longrightarrow\quad x \in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ with strict inequalities at both ends.

4. Combine all the restrictions
We must honour simultaneously:
• $$0 \le x \le 4$$ (from the inverse-cosine term), and
• $$x \in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ with open ends (from the logarithmic term).

Intersecting these two intervals gives
$$x \in [0, \frac{\pi}{2})$$ because:
• The lower endpoint 0 is allowed (cos 0 = 1 > 0 and the arccos argument equals -1, which is permitted).
• The upper endpoint $$\frac{\pi}{2}$$ is excluded because $$\cos\!\left(\frac{\pi}{2}\right)=0$$ makes the logarithm undefined.

This is the largest interval contained in $$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ on which every part of the function is defined.

Hence the required interval is $$\left[0, \frac{\pi}{2}\right).$$

Option D which is: $$\left[0, \frac{\pi}{2}\right)$$