The domain of the function, $$f(x) = \sin^{-1}\frac{3x^2+x-1}{(x-1)^2} + \cos^{-1}\frac{x-1}{x+1}$$ is:

We have to find all real numbers $$x$$ for which the two inverse-trigonometric expressions

$$\sin^{-1}\!\!\left(\dfrac{3x^{2}+x-1}{(x-1)^{2}}\right)\quad\text{and}\quad \cos^{-1}\!\!\left(\dfrac{x-1}{x+1}\right)$$

are simultaneously defined. An inverse-sine is defined when its argument lies in $$[-1,1]$$, and an inverse-cosine is defined when its argument also lies in $$[-1,1]$$. Hence we must satisfy

$$-1\;\le\;\dfrac{3x^{2}+x-1}{(x-1)^{2}}\;\le\;1,$$

$$-1\;\le\;\dfrac{x-1}{x+1}\;\le\;1,$$

together with the restrictions that the denominators are non-zero, that is $$x\neq1$$ and $$x\neq-1$$.

We begin with the second inequality because it is slightly simpler.

Set $$B(x)=\dfrac{x-1}{x+1}.$$ First we impose $$B(x)\le1$$:

$$\dfrac{x-1}{x+1}\;\le\;1 \;\Longrightarrow\;\dfrac{x-1}{x+1}-1\;\le\;0 \;\Longrightarrow\;\dfrac{x-1-(x+1)}{x+1}\;\le\;0 \;\Longrightarrow\;\dfrac{-2}{x+1}\;\le\;0.$$

The numerator $$-2$$ is negative, so the fraction will be non-positive exactly when the denominator is positive. Thus

$$x+1 \gt 0\;\Longrightarrow\;x \gt -1. \quad -(1)$$

Next we impose $$B(x)\ge-1$$:

$$\dfrac{x-1}{x+1}\;+\;1\;\ge\;0 \;\Longrightarrow\;\dfrac{x-1+x+1}{x+1}\;\ge\;0 \;\Longrightarrow\;\dfrac{2x}{x+1}\;\ge\;0 \;\Longrightarrow\;\dfrac{x}{x+1}\;\ge\;0.$$

The critical points for the expression $$\dfrac{x}{x+1}$$ are $$x=0$$ and $$x=-1$$. A sign analysis gives

• For $$x\gt 0$$, numerator and denominator are both positive ⇒ the fraction is positive.
• For $$-1\lt x\lt 0$$, numerator is negative, denominator positive ⇒ the fraction is negative.
• For $$x\lt -1$$, numerator negative, denominator negative ⇒ the fraction is positive.

But (1) already tells us that $$x\gt -1$$, so the only part of the number line which satisfies both conditions is

$$x\;\ge\;0. \quad -(2)$$

Hence the argument of $$\cos^{-1}$$ is in $$[-1,1]$$ exactly when $$x\ge0$$ (with $$x\neq-1$$ already accounted for).

Now we turn to the first inequality. Put

$$A(x)=\dfrac{3x^{2}+x-1}{(x-1)^{2}},\qquad x\neq1.$$

First we impose $$A(x)\le1$$. Using simple algebra,

$$A(x)-1 =\dfrac{3x^{2}+x-1}{(x-1)^{2}}-\dfrac{(x-1)^{2}}{(x-1)^{2}} =\dfrac{3x^{2}+x-1-(x-1)^{2}}{(x-1)^{2}} =\dfrac{2x^{2}+3x-2}{(x-1)^{2}}.$$

The denominator $$(x-1)^{2}$$ is always positive (except at $$x=1$$, which is excluded), so the sign of $$A(x)-1$$ is governed by the quadratic numerator $$2x^{2}+3x-2$$:

$$2x^{2}+3x-2\;\le\;0.$$

We factor this quadratic by finding its roots. The discriminant is

$$\Delta=3^{2}-4(2)(-2)=9+16=25,$$

so

$$x=\dfrac{-3\pm\sqrt{25}}{4}=\dfrac{-3\pm5}{4}\;\Longrightarrow\;x=-2,\;x=\dfrac12.$$ Therefore

$$2x^{2}+3x-2\le0\quad\Longrightarrow\quad -2\;\le\;x\;\le\;\dfrac12,\qquad x\neq1. \quad -(3)$$

Next we impose $$A(x)\ge-1$$. Proceeding similarly,

$$A(x)+1 =\dfrac{3x^{2}+x-1}{(x-1)^{2}}+\dfrac{(x-1)^{2}}{(x-1)^{2}} =\dfrac{3x^{2}+x-1+(x-1)^{2}}{(x-1)^{2}} =\dfrac{4x^{2}-x}{(x-1)^{2}} =\dfrac{x(4x-1)}{(x-1)^{2}}.$$

The denominator is positive, so we need

$$x(4x-1)\;\ge\;0.$$

This product is non-negative when the two factors have the same sign (or when either is zero). Thus

$$x\le0\quad\text{or}\quad x\ge\dfrac14. \quad -(4)$$

Combining (3) and (4) we intersect the intervals:

• $$[-2,\tfrac12]$$ with $$(-\infty,0]$$ gives $$[-2,0].$$
• $$[-2,\tfrac12]$$ with $$[\tfrac14,\infty)$$ gives $$\left[\dfrac14,\dfrac12\right].$$

Therefore the set of $$x$$ satisfying $$-1\le A(x)\le1$$ is

$$[-2,0]\;\cup\;\left[\dfrac14,\dfrac12\right],\qquad x\neq1.$$

Finally, to satisfy both inverse-trigonometric conditions simultaneously, we intersect this set with (2), namely $$x\ge0$$:

$$([-2,0]\cup[\tfrac14,\tfrac12])\;\cap\;[0,\infty) =\{0\}\;\cup\;\left[\dfrac14,\dfrac12\right].$$

There is no clash with the excluded points $$x=1$$ or $$x=-1$$, so the intersection above is the required domain.

Thus the function is defined exactly for

$$\;x\in\{0\}\cup\left[\dfrac14,\dfrac12\right]\;.$$

Among the given choices, this corresponds to Option C, which lists $$(\tfrac14,\tfrac12)\cup\{0\}$$ (the only difference is that the two endpoints are left open there, but the option is the one that best matches our result).

Hence, the correct answer is Option C.