Domain and Range of ITF Questions in English

(B) The function $f(x) = \sin^{-1}x$ is the inverse of the sine function $y = \sin x$,where the domain of $\sin x$ is restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to make it bijective.
By the definition of inverse functions,the domain of the inverse function is the range of the original function,and the range of the inverse function is the domain of the original function.
Since the range of the sine function $y = \sin x$ is $[-1, 1]$,the domain of its inverse function $\sin^{-1}x$ is $[-1, 1]$.

(B) The function $f(x) = \tan^{-1} x$ is defined for all real numbers $x \in \mathbb{R}$.
By definition,the principal value branch of the inverse tangent function is the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Therefore,the range of $\tan^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Thus,the correct option is $B$.

(B) The domain of the function $f(x) = \sin^{-1}(u)$ is defined by the condition $-1 \le u \le 1$.
For the given function $f(x) = \sin^{-1}(5x)$,we set $u = 5x$.
Thus,the condition becomes $-1 \le 5x \le 1$.
Dividing the entire inequality by $5$,we get $-\frac{1}{5} \le x \le \frac{1}{5}$.
Therefore,the domain of the function is $[-\frac{1}{5}, \frac{1}{5}]$.

(A) The given function is $f(x) = \sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$.
For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \le u \le 1$.
Therefore,we must have $-1 \le \log_3 \left( \frac{x}{3} \right) \le 1$.
Applying the definition of the logarithm,we exponentiate with base $3$:
$3^{-1} \le \frac{x}{3} \le 3^1$.
This simplifies to $\frac{1}{3} \le \frac{x}{3} \le 3$.
Multiplying the entire inequality by $3$,we get $1 \le x \le 9$.
Thus,the domain is $x \in [1, 9]$.

(A) The function is defined as $f(x) = \sin^{-1}[\log_2(x/2)]$.
Since the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$,we must have:
$-1 \le \log_2(x/2) \le 1$.
Applying the definition of the logarithm,we exponentiate with base $2$:
$2^{-1} \le x/2 \le 2^1$.
This simplifies to:
$1/2 \le x/2 \le 2$.
Multiplying the entire inequality by $2$,we get:
$1 \le x \le 4$.
Therefore,the domain of the function is $x \in [1, 4]$.

(C) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.
Therefore,we must have $-1 \le 1 + 3x + 2x^2 \le 1$.
Case $I$: $2x^2 + 3x + 1 \ge -1$
$2x^2 + 3x + 2 \ge 0$.
The discriminant $D = 3^2 - 4(2)(2) = 9 - 16 = -7 < 0$. Since the coefficient of $x^2$ is positive,the expression $2x^2 + 3x + 2$ is always positive for all $x \in \mathbb{R}$.
Case $II$: $2x^2 + 3x + 1 \le 1$
$2x^2 + 3x \le 0$
$x(2x + 3) \le 0$.
The roots are $x = 0$ and $x = -\frac{3}{2}$.
Testing the intervals,the inequality holds for $x \in \left[ -\frac{3}{2}, 0 \right]$.
Combining both cases,the domain is $\left[ -\frac{3}{2}, 0 \right]$.

(B) To define the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$,we must satisfy two conditions:
$1$. The argument of the square root in the denominator must be strictly positive:
$9 - x^2 > 0$
$x^2 < 9$
$-3 < x < 3$ --- $(i)$
$2$. The argument of the inverse sine function must lie in the interval $[-1, 1]$:
$-1 \le x - 3 \le 1$
Adding $3$ to all parts:
$2 \le x \le 4$ --- $(ii)$
Taking the intersection of conditions $(i)$ and $(ii)$:
$x \in (-3, 3) \cap [2, 4]$
$x \in [2, 3)$
Thus,the domain of the function is $[2, 3)$.

(B) For the function $f(x) = \sin^{-1}(\sqrt{x})$ to be defined,the argument of the inverse sine function must lie in the interval $[-1, 1]$.
Thus,we must have $-1 \le \sqrt{x} \le 1$.
Since $\sqrt{x}$ is always non-negative,the condition $-1 \le \sqrt{x}$ is always satisfied for all $x \ge 0$.
Now,we solve $\sqrt{x} \le 1$.
Squaring both sides,we get $x \le 1^2$,which implies $x \le 1$.
Also,for $\sqrt{x}$ to be defined,we must have $x \ge 0$.
Combining these conditions,we get $0 \le x \le 1$.
Therefore,the function is defined in the interval $[0, 1]$.

(B) For $-1 < x < 1$,we use the substitution $x = \tan \theta$,where $\theta \in (-\frac{\pi}{4}, \frac{\pi}{4})$.
Then,$\frac{2x}{1-x^2} = \frac{2\tan \theta}{1-\tan^2 \theta} = \tan(2\theta)$.
Thus,$f(x) = \tan^{-1}(\tan(2\theta)) = 2\theta = 2\tan^{-1}x$.
Since $x \in (-1, 1)$,$\tan^{-1}x \in (-\frac{\pi}{4}, \frac{\pi}{4})$.
Therefore,$f(x) \in (2 \times -\frac{\pi}{4}, 2 \times \frac{\pi}{4}) = (-\frac{\pi}{2}, \frac{\pi}{2})$.
For the function to be onto,the codomain $B$ must be equal to the range of the function.
Hence,$B = (-\frac{\pi}{2}, \frac{\pi}{2})$.

(D) The domain of the function $f(x) = \sin^{-1}(\log_3 x)$ is determined by the condition that the argument of the inverse sine function must lie in the interval $[-1, 1]$.
Thus,we have the inequality: $-1 \le \log_3 x \le 1$.
Since the base of the logarithm is $3$ (which is greater than $1$),we can exponentiate each part of the inequality with base $3$ without changing the direction of the inequality signs:
$3^{-1} \le x \le 3^1$.
Simplifying the expressions,we get:
$\frac{1}{3} \le x \le 3$.
Therefore,the domain of the function is $[\frac{1}{3}, 3]$.

(A) For the function $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$ to be defined,the arguments of $\sin^{-1}$ and $\cos^{-1}$ must lie in the interval $[-1, 1]$.
$1$. For $\sin^{-1} \left( \frac{1 - |x|}{3} \right)$,we have:
$-1 \le \frac{1 - |x|}{3} \le 1$
$-3 \le 1 - |x| \le 3$
$-4 \le -|x| \le 2$
$-2 \le |x| \le 4$
Since $|x| \ge 0$,this implies $0 \le |x| \le 4$.
$2$. For $\cos^{-1} \left( \frac{|x| - 3}{5} \right)$,we have:
$-1 \le \frac{|x| - 3}{5} \le 1$
$-5 \le |x| - 3 \le 5$
$-2 \le |x| \le 8$
Since $|x| \ge 0$,this implies $0 \le |x| \le 8$.
Taking the intersection of both conditions:
$0 \le |x| \le 4$
This inequality is equivalent to $-4 \le x \le 4$.
Thus,the domain of $f(x)$ is $[-4, 4]$.

(D) For the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ to be defined,the expression inside the square root must be strictly positive:
$\ln(\cot^{-1}x) > 0$
Since the base of the logarithm is $e > 1$,we have:
$\cot^{-1}x > e^0$
$\cot^{-1}x > 1$
Since the function $y = \cot^{-1}x$ is a strictly decreasing function,the inequality reverses when we apply the cotangent function:
$x < \cot(1)$
Thus,the domain of the function is $( -\infty, \cot 1 )$.

(B) Let $g(x) = 5x^2 - 8x + 4$.
Completing the square,$g(x) = 5(x^2 - \frac{8}{5}x) + 4 = 5(x - \frac{4}{5})^2 + 4 - 5(\frac{16}{25}) = 5(x - \frac{4}{5})^2 + \frac{4}{5}$.
Since $5(x - \frac{4}{5})^2 \geq 0$,the range of $g(x)$ is $[\frac{4}{5}, \infty)$.
Let $u = \log_{4/5}(g(x))$. Since the base $4/5 < 1$,the function $\log_{4/5}(t)$ is strictly decreasing.
For $t \in [\frac{4}{5}, \infty)$,$\log_{4/5}(t) \in (\log_{4/5}(\infty), \log_{4/5}(\frac{4}{5})] = (-\infty, 1]$.
Now,$f(x) = \cot^{-1}(u)$ where $u \in (-\infty, 1]$.
Since $\cot^{-1}(u)$ is a strictly decreasing function,the range is $[\cot^{-1}(1), \cot^{-1}(-\infty)) = [\frac{\pi}{4}, \pi)$.

(C) For the function $f(x) = \sqrt{2 - \sec^{-1}x}$ to be defined,the expression inside the square root must be non-negative:
$2 - \sec^{-1}x \ge 0$
$\sec^{-1}x \le 2$
Also,the domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$.
Since $\sec^{-1}x \le 2$,and the range of $\sec^{-1}x$ is $[0, \pi] \setminus \{\pi/2\}$,we consider the inequality $\sec^{-1}x \le 2$.
For $x \ge 1$,$\sec^{-1}x \in [0, \pi/2)$,so $\sec^{-1}x \le 2$ is always true for all $x \in [1, \infty)$.
For $x \le -1$,$\sec^{-1}x \in (\pi/2, \pi]$. We need $\sec^{-1}x \le 2$. Since $2 < \pi \approx 3.14$,this holds for values where $\sec^{-1}x \le 2$.
Taking the secant of both sides (noting that $\sec$ is increasing on $[0, \pi/2)$ and decreasing on $(\pi/2, \pi]$ is not directly applicable,but we know $\sec^{-1}x \le 2 \implies x \le \sec 2$ for $x \le -1$ because $\sec 2$ is negative).
Thus,the domain is $(-\infty, \sec 2] \cup [1, \infty)$.

(D) Let $f(x) = \sin^{-1}\left(\frac{1+x^2}{2+x^2}\right)$.
We can rewrite the argument as $\frac{1+x^2}{2+x^2} = \frac{2+x^2-1}{2+x^2} = 1 - \frac{1}{2+x^2}$.
Since $x^2 \ge 0$,we have $2+x^2 \ge 2$,which implies $0 < \frac{1}{2+x^2} \le \frac{1}{2}$.
Multiplying by $-1$,we get $-\frac{1}{2} \le -\frac{1}{2+x^2} < 0$.
Adding $1$ to all parts,we get $1 - \frac{1}{2} \le 1 - \frac{1}{2+x^2} < 1$,which simplifies to $\frac{1}{2} \le \frac{1+x^2}{2+x^2} < 1$.
Since the function $\sin^{-1}(u)$ is strictly increasing,the range is $[\sin^{-1}(\frac{1}{2}), \sin^{-1}(1))$.
Therefore,the range is $[\frac{\pi}{6}, \frac{\pi}{2})$.

(D) Let $g(x) = x^2 + x + 1$.
Since $x^2 + x + 1 = (x + 1/2)^2 + 3/4$,the minimum value of $g(x)$ is $3/4$ at $x = -1/2$.
As $x \to \pm \infty$,$g(x) \to \infty$. However,for $\sin^{-1}(\sqrt{g(x)})$ to be defined,we must have $0 \le \sqrt{g(x)} \le 1$,which implies $0 \le g(x) \le 1$.
Thus,$3/4 \le x^2 + x + 1 \le 1$.
Taking the square root,we get $\sqrt{3}/2 \le \sqrt{x^2 + x + 1} \le 1$.
Applying the $\sin^{-1}$ function,we get $\sin^{-1}(\sqrt{3}/2) \le \sin^{-1}(\sqrt{x^2 + x + 1}) \le \sin^{-1}(1)$.
This results in $\pi/3 \le f(x) \le \pi/2$.
Therefore,the range is $\left[ \frac{\pi}{3}, \frac{\pi}{2} \right]$.

(C) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive:
$\log_{\frac{\pi}{4}}(\sin^{-1} x) - 1 > 0$
$\Rightarrow \log_{\frac{\pi}{4}}(\sin^{-1} x) > 1$
Since the base $\frac{\pi}{4} < 1$,the inequality sign reverses when we remove the logarithm:
$\sin^{-1} x < (\frac{\pi}{4})^1 = \frac{\pi}{4}$
Also,the argument of the logarithm must be positive:
$\sin^{-1} x > 0 \Rightarrow x > 0$
Additionally,for $\sin^{-1} x$ to be defined,$x \in [-1, 1]$.
Combining these conditions:
$0 < \sin^{-1} x < \frac{\pi}{4}$
Taking the sine of all parts:
$\sin(0) < x < \sin(\frac{\pi}{4})$
$0 < x < \frac{1}{\sqrt{2}}$
Thus,the domain is $x \in \left( 0, \frac{1}{\sqrt{2}} \right)$.

(C) The function $f(x)$ is defined if the arguments of $\sin^{-1}$,$\cos^{-1}$,and $\tan^{-1}$ satisfy their respective domain conditions.
For $\sin^{-1}(u)$ and $\cos^{-1}(u)$,we must have $-1 \leq u \leq 1$.
For $\tan^{-1}(u)$,$u$ can be any real number.
Thus,we require $-1 \leq \frac{2-|x|}{4} \leq 1$.
Multiplying by $4$,we get $-4 \leq 2-|x| \leq 4$.
Subtracting $2$ from all parts,we get $-6 \leq -|x| \leq 2$.
Multiplying by $-1$ (and reversing the inequalities),we get $-2 \leq |x| \leq 6$.
Since $|x|$ is always non-negative,the condition $-2 \leq |x|$ is always true for all $x \in R$.
Therefore,we only need $|x| \leq 6$,which implies $-6 \leq x \leq 6$.
Hence,the domain is $x \in [-6, 6]$.

(A) For the function $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.
Since $|x|+5 > 0$ and $x^2+1 > 0$ for all $x \in \mathbb{R}$,the condition $\frac{|x|+5}{x^2+1} \geq -1$ is always satisfied.
Thus,we only need to solve $\frac{|x|+5}{x^2+1} \leq 1$.
Multiplying by $x^2+1$ (which is positive),we get $|x|+5 \leq x^2+1$.
Rearranging the terms,we have $x^2 - |x| - 4 \geq 0$.
Let $t = |x|$,where $t \geq 0$. Then $t^2 - t - 4 \geq 0$.
The roots of $t^2 - t - 4 = 0$ are $t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$.
Since $t = |x| \geq 0$,we must have $t \geq \frac{1+\sqrt{17}}{2}$.
Therefore,$|x| \geq \frac{1+\sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1+\sqrt{17}}{2}\right] \cup \left[\frac{1+\sqrt{17}}{2}, \infty\right)$.
Comparing this with the given domain $(-\infty, -a] \cup [a, \infty)$,we find $a = \frac{1+\sqrt{17}}{2}$.

(D) The domain of $\operatorname{cosec}^{-1}(y)$ is $y \in (-\infty, -1] \cup [1, \infty)$.
For the function $\operatorname{cosec}^{-1}\left(\frac{1+x}{x}\right)$,we must have $\frac{1+x}{x} \in (-\infty, -1] \cup [1, \infty)$.
Case $1$: $\frac{1+x}{x} \geq 1$
$\frac{1}{x} + 1 \geq 1 \implies \frac{1}{x} \geq 0 \implies x > 0$.
Case $2$: $\frac{1+x}{x} \leq -1$
$\frac{1}{x} + 1 \leq -1 \implies \frac{1}{x} \leq -2$.
Since $\frac{1}{x} \leq -2$,$x$ must be negative. Multiplying by $x$ (which is negative) reverses the inequality:
$1 \geq -2x \implies x \geq -\frac{1}{2}$.
Thus,$x \in [-\frac{1}{2}, 0)$.
Combining both cases,the domain is $x \in [-\frac{1}{2}, 0) \cup (0, \infty)$,which can be written as $[-\frac{1}{2}, \infty) - \{0\}$.

(C) For the function $f(x) = \sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$ to be defined,both parts must be defined.
$1$. For $\cos^{-1}\left(\frac{x-1}{x+1}\right)$,we require $-1 \leq \frac{x-1}{x+1} \leq 1$.
Solving $\frac{x-1}{x+1} \leq 1 \Rightarrow \frac{x-1-x-1}{x+1} \leq 0 \Rightarrow \frac{-2}{x+1} \leq 0 \Rightarrow x+1 > 0 \Rightarrow x > -1$.
Solving $\frac{x-1}{x+1} \geq -1 \Rightarrow \frac{x-1+x+1}{x+1} \geq 0 \Rightarrow \frac{2x}{x+1} \geq 0 \Rightarrow x \in (-\infty, -1) \cup [0, \infty)$.
Combining these,we get $x \in [0, \infty)$.
$2$. For $\sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right)$,we require $-1 \leq \frac{3x^2+x-1}{(x-1)^2} \leq 1$.
Solving $\frac{3x^2+x-1}{(x-1)^2} \leq 1 \Rightarrow 3x^2+x-1 \leq x^2-2x+1 \Rightarrow 2x^2+3x-2 \leq 0 \Rightarrow (2x-1)(x+2) \leq 0 \Rightarrow x \in [-2, 1/2]$.
Solving $\frac{3x^2+x-1}{(x-1)^2} \geq -1 \Rightarrow 3x^2+x-1 \geq -x^2+2x-1 \Rightarrow 4x^2-x \geq 0 \Rightarrow x(4x-1) \geq 0 \Rightarrow x \in (-\infty, 0] \cup [1/4, \infty)$.
Combining these,we get $x \in [-2, 0] \cup [1/4, 1/2]$.
Taking the intersection of the domains from $(1)$ and $(2)$,we get $x \in [0, \infty) \cap ([-2, 0] \cup [1/4, 1/2]) = \{0\} \cup [1/4, 1/2]$.

(A) The given equation is $\tan ^{-1} \sqrt{x^{2}+x}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$.
For the equation to be defined,the arguments of the inverse trigonometric functions must satisfy their domain conditions.
$1$. For $\tan ^{-1} \sqrt{x^{2}+x}$,we require $x^{2}+x \geq 0$.
$2$. For $\sin ^{-1} \sqrt{x^{2}+x+1}$,we require $0 \leq \sqrt{x^{2}+x+1} \leq 1$,which implies $0 \leq x^{2}+x+1 \leq 1$.
Since $x^{2}+x+1 = (x+\frac{1}{2})^{2} + \frac{3}{4}$,the minimum value of $x^{2}+x+1$ is $\frac{3}{4}$.
Thus,the condition $x^{2}+x+1 \leq 1$ implies $x^{2}+x+1 \leq 1$,or $x^{2}+x \leq 0$.
Combining $x^{2}+x \geq 0$ and $x^{2}+x \leq 0$,we must have $x^{2}+x = 0$.
This gives $x(x+1) = 0$,so $x=0$ or $x=-1$.
If $x=0$,the equation becomes $\tan ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \neq \frac{\pi}{4}$.
If $x=-1$,the equation becomes $\tan ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \neq \frac{\pi}{4}$.
Since neither value satisfies the equation,the number of real roots is $0$.

(B) For the function $f(x)$ to be defined,the following conditions must be satisfied:
$1$. The argument of $\cos^{-1}$ must be in $[0, 1]$: $0 \leq \sqrt{x^2-x+1} \leq 1$.
Squaring gives $0 \leq x^2-x+1 \leq 1$.
$x^2-x+1 \leq 1 \Rightarrow x^2-x \leq 0 \Rightarrow x(x-1) \leq 0 \Rightarrow x \in [0, 1]$.
$2$. The denominator must be non-zero and the argument of the square root must be positive: $\sin^{-1}(\frac{2x-1}{2}) > 0$.
Since the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we also require $\frac{2x-1}{2} \leq 1$.
So,$0 < \frac{2x-1}{2} \leq 1$.
$0 < 2x-1 \leq 2$.
$1 < 2x \leq 3$.
$\frac{1}{2} < x \leq \frac{3}{2}$.
Taking the intersection of $x \in [0, 1]$ and $x \in (\frac{1}{2}, \frac{3}{2}]$,we get $x \in (\frac{1}{2}, 1]$.
Thus,$\alpha = \frac{1}{2}$ and $\beta = 1$.
Therefore,$\alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2}$.

(D) For the function $\cos^{-1}(u)$ to be defined,we must have $-1 \leq u \leq 1$.
Thus,$-1 \leq \frac{2 \sin^{-1}(\frac{1}{4x^2-1})}{\pi} \leq 1$.
Multiplying by $\frac{\pi}{2}$,we get $-\frac{\pi}{2} \leq \sin^{-1}(\frac{1}{4x^2-1}) \leq \frac{\pi}{2}$.
Since the range of $\sin^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,this inequality is always satisfied for all $y$ in the domain of $\sin^{-1}$,which is $[-1, 1]$.
Therefore,we need $-1 \leq \frac{1}{4x^2-1} \leq 1$.
Case $1$: $\frac{1}{4x^2-1} \leq 1 \implies \frac{1 - (4x^2-1)}{4x^2-1} \leq 0 \implies \frac{2-4x^2}{4x^2-1} \leq 0 \implies \frac{2(1-2x^2)}{(2x-1)(2x+1)} \geq 0$.
This holds for $x \in (-\infty, -\frac{1}{\sqrt{2}}] \cup (-\frac{1}{2}, \frac{1}{2}) \cup [\frac{1}{\sqrt{2}}, \infty)$.
Case $2$: $\frac{1}{4x^2-1} \geq -1 \implies \frac{1 + 4x^2 - 1}{4x^2-1} \geq 0 \implies \frac{4x^2}{4x^2-1} \geq 0$.
This holds for $x \in (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty) \cup \{0\}$.
Taking the intersection of both cases,we get $x \in (-\infty, -\frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, \infty) \cup \{0\}$.

(B) The domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.
Thus,we require $\left|\frac{x^{2}-4x+2}{x^{2}+3}\right| \leq 1$.
This implies $-1 \leq \frac{x^{2}-4x+2}{x^{2}+3} \leq 1$.
Since $x^{2}+3 > 0$ for all $x \in \mathbb{R}$,we can multiply by $(x^{2}+3)$:
$-(x^{2}+3) \leq x^{2}-4x+2 \leq x^{2}+3$.
First inequality: $-x^{2}-3 \leq x^{2}-4x+2 \implies 2x^{2}-4x+5 \geq 0$. The discriminant $D = (-4)^{2} - 4(2)(5) = 16 - 40 = -24 < 0$. Since the leading coefficient is positive,$2x^{2}-4x+5 > 0$ for all $x \in \mathbb{R}$.
Second inequality: $x^{2}-4x+2 \leq x^{2}+3 \implies -4x \leq 1 \implies x \geq -\frac{1}{4}$.
Therefore,the domain is $[-\frac{1}{4}, \infty)$.

(C) For the function $f(x) = \sin^{-1}(g(x))$ to be defined,the argument $g(x)$ must satisfy $-1 \leq g(x) \leq 1$.
Step $1$: Solve $\frac{x^{2}-3x+2}{x^{2}+2x+7} \geq -1$.
$x^{2}-3x+2 \geq -(x^{2}+2x+7)$
$2x^{2}-x+9 \geq 0$.
The discriminant $D = (-1)^{2} - 4(2)(9) = 1 - 72 = -71 < 0$. Since the coefficient of $x^{2}$ is positive,$2x^{2}-x+9$ is always positive for all $x \in \mathbb{R}$.
Step $2$: Solve $\frac{x^{2}-3x+2}{x^{2}+2x+7} \leq 1$.
$x^{2}-3x+2 \leq x^{2}+2x+7$
$-3x+2 \leq 2x+7$
$-5 \leq 5x \Rightarrow x \geq -1$.
Combining both conditions,the domain is $x \in [-1, \infty)$.

(C) The function $f(x) = \sec^{-1}\left(\frac{2x}{5x+3}\right)$ is defined when $\left|\frac{2x}{5x+3}\right| \geq 1$ and $5x+3 \neq 0$.
This implies $\left|\frac{2x}{5x+3}\right| \geq 1$,which means $(2x)^2 \geq (5x+3)^2$.
$(2x)^2 - (5x+3)^2 \geq 0$
$(2x - 5x - 3)(2x + 5x + 3) \geq 0$
$(-3x - 3)(7x + 3) \geq 0$
$-(3x + 3)(7x + 3) \geq 0 \Rightarrow (x + 1)(7x + 3) \leq 0$.
The solution to this inequality is $x \in [-1, -3/7]$.
Additionally,the denominator $5x+3 \neq 0$ implies $x \neq -3/5$.
Thus,the domain is $[-1, -3/5) \cup (-3/5, -3/7]$.
Comparing this with $[\alpha, \beta) \cup (\gamma, \delta]$,we get $\alpha = -1, \beta = -3/5, \gamma = -3/5, \delta = -3/7$.
Now,calculate $|3\alpha + 10(\beta + \gamma) + 21\delta| = |3(-1) + 10(-3/5 - 3/5) + 21(-3/7)|$.
$= |-3 + 10(-6/5) + 3(-3)| = |-3 - 12 - 9| = |-24| = 24$.

(B) Let $u = \frac{x^2}{x^2+1}$.
Since $x^2 \ge 0$,we have $x^2+1 \ge 1$,so $0 \le \frac{x^2}{x^2+1} < 1$.
Thus,the range of $u$ is $[0, 1)$.
Now,$f(x) = 4 \sin^{-1}(u)$.
Since $u \in [0, 1)$,$\sin^{-1}(u) \in [\sin^{-1}(0), \sin^{-1}(1)) = [0, \frac{\pi}{2})$.
Multiplying by $4$,we get $f(x) \in [4 \times 0, 4 \times \frac{\pi}{2}) = [0, 2\pi)$.
Therefore,the range of $f(x)$ is $[0, 2\pi)$.

(D) The domain of $f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$ requires that the argument of $\sin^{-1}$ lies in the interval $[-1, 1]$ and the denominator is non-zero.
$1$. Denominator condition: $2x + 3 \neq 0 \implies x \neq -\frac{3}{2}$.
$2$. Inequality condition: $\left|\frac{x-1}{2x+3}\right| \leq 1$.
Since $2x+3 \neq 0$,we have $|x-1| \leq |2x+3|$.
Squaring both sides: $(x-1)^2 \leq (2x+3)^2$.
$x^2 - 2x + 1 \leq 4x^2 + 12x + 9$.
$3x^2 + 14x + 8 \geq 0$.
Factoring the quadratic: $(3x + 2)(x + 4) \geq 0$.
This inequality holds for $x \in (-\infty, -4] \cup [-\frac{2}{3}, \infty)$.
$3$. Combining with the condition $x \neq -\frac{3}{2}$:
The domain is $(-\infty, -4] \cup [-\frac{2}{3}, \infty) \setminus \{-\frac{3}{2}\}$.
However,the problem states the domain is $R - (\alpha, \beta)$. This implies the excluded region is an open interval. Looking at the complement,the excluded values are $(-4, -\frac{2}{3})$.
Thus,$\alpha = -4$ and $\beta = -\frac{2}{3}$.
$4$. Calculating $12\alpha\beta$:
$12 \times (-4) \times (-\frac{2}{3}) = 12 \times \frac{8}{3} = 4 \times 8 = 32$.

(B) The function $f(x) = \sec^{-1}(u)$ is defined for $|u| \geq 1$,which means $u \leq -1$ or $u \geq 1$.
Here,$u = 2[x] + 1$.
So,$2[x] + 1 \leq -1$ or $2[x] + 1 \geq 1$.
Case $1$: $2[x] + 1 \leq -1 \Rightarrow 2[x] \leq -2 \Rightarrow [x] \leq -1$. This implies $x < 0$.
Case $2$: $2[x] + 1 \geq 1 \Rightarrow 2[x] \geq 0 \Rightarrow [x] \geq 0$. This implies $x \geq 0$.
Combining both cases,$x \in (-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$.

(A) For the function $f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}}$ to be defined,the expression inside the square root must be non-negative:
$\sin^{-1}(2x) + \frac{\pi}{6} \geq 0$
$\sin^{-1}(2x) \geq -\frac{\pi}{6}$
Also,the domain of $\sin^{-1}(\theta)$ is $[-1, 1]$,so $-1 \leq 2x \leq 1$,which implies $-0.5 \leq x \leq 0.5$.
The range of $\sin^{-1}(2x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Combining these,we have $-\frac{\pi}{6} \leq \sin^{-1}(2x) \leq \frac{\pi}{2}$.
Taking the sine of all parts:
$\sin(-\frac{\pi}{6}) \leq 2x \leq \sin(\frac{\pi}{2})$
$-\frac{1}{2} \leq 2x \leq 1$
Dividing by $2$:
$-\frac{1}{4} \leq x \leq \frac{1}{2}$
Thus,the domain is $\left[-\frac{1}{4}, \frac{1}{2}\right]$.

(B) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive and the argument of $\sin^{-1}$ must be in the interval $[-1, 1]$.
$1$. For the denominator: $9 - x^2 > 0$ $\Rightarrow x^2 < 9$ $\Rightarrow -3 < x < 3 \dots (i)$.
$2$. For the numerator: $-1 \leq x - 3 \leq 1 \Rightarrow 2 \leq x \leq 4 \dots (ii)$.
$3$. Taking the intersection of $(i)$ and $(ii)$,we get $2 \leq x < 3$.
Thus,the domain is $[2, 3)$.

(C) $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is defined if $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.
Since $|x|+5 > 0$ and $x^2+1 > 0$,the left inequality $\frac{|x|+5}{x^2+1} \geq -1$ is always true.
We only need to solve $\frac{|x|+5}{x^2+1} \leq 1$.
$|x|+5 \leq x^2+1$
$x^2 - |x| - 4 \geq 0$.
Let $t = |x|$,where $t \geq 0$. Then $t^2 - t - 4 \geq 0$.
The roots of $t^2 - t - 4 = 0$ are $t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$.
Since $t \geq 0$,we have $t \geq \frac{1 + \sqrt{17}}{2}$.
Thus,$|x| \geq \frac{1 + \sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1 + \sqrt{17}}{2}\right] \cup \left[\frac{1 + \sqrt{17}}{2}, \infty\right)$.
Comparing this with $(-\infty, -a] \cup [a, \infty)$,we get $a = \frac{1 + \sqrt{17}}{2}$.

(C) The function $y = \tan^{-1} x$ represents the inverse tangent function.
By definition,the principal value branch of the inverse tangent function is defined as the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Therefore,for any real number $x$,the range of $y = \tan^{-1} x$ is $-\frac{\pi}{2} < y < \frac{\pi}{2}$.

(D) The principal value branch of the inverse cosine function,denoted by $\cos^{-1} x$,is defined as the interval $[0, \pi]$.
Therefore,if $\cos^{-1} x = y$,then the range of $y$ must be $0 \leq y \leq \pi$.
Thus,the correct option is $D$.

(D) The function $f(x) = \sin^{-1}(x)$ is the inverse of the sine function $f(x) = \sin(x)$.
For the sine function,the range is $[-1, 1]$.
By the definition of inverse trigonometric functions,the domain of the inverse function is equal to the range of the original function.
Therefore,the domain of $\sin^{-1}(x)$ is $[-1, 1]$.

(A) We know that the domain of $\cos^{-1} x$ is $[-1, 1]$.
For $x \in [-1, 1]$,the range of $\cos^{-1} x$ is $[0, \pi]$.
Since $\cos^{-1} x$ takes all values in the interval $[0, \pi]$,the absolute value $\left|\cos^{-1} x\right|$ will also take all values in the interval $[0, \pi]$ because all values in $[0, \pi]$ are non-negative.
Therefore,the range of $\left|\cos^{-1} x\right|$ is $[0, \pi]$.

(D) We know that the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$.
Here,the argument is $\frac{\pi}{3}$.
Since $\pi \approx 3.14159$,we have $\frac{\pi}{3} \approx 1.047$.
Because $1.047 > 1$,the value $\frac{\pi}{3}$ lies outside the domain $[-1, 1]$.
Therefore,$\sin ^{-1} \left(\frac{\pi}{3}\right)$ and $\cos ^{-1} \left(\frac{\pi}{3}\right)$ are not defined in the set of real numbers.
Thus,the expression $\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)$ does not exist.

(D) The function $f(x) = \sec^{-1} x$ is the inverse of the secant function restricted to the domain $[0, \pi] - \{\frac{\pi}{2}\}$.
By definition,the range of the principal value branch of $\sec^{-1} x$ is the set of all values in the interval $[0, \pi]$ excluding $\frac{\pi}{2}$,because $\sec x$ is undefined at $\frac{\pi}{2}$.
Therefore,the range is $[0, \pi] - \{\frac{\pi}{2}\}$.

(C) The given function is $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$.
For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \leq u \leq 1$.
Therefore,we must have $-1 \leq \log_{2}\left(\frac{x}{2}\right) \leq 1$.
Applying the definition of the logarithm,this inequality is equivalent to $2^{-1} \leq \frac{x}{2} \leq 2^{1}$.
This simplifies to $\frac{1}{2} \leq \frac{x}{2} \leq 2$.
Multiplying the entire inequality by $2$,we get $1 \leq x \leq 4$.
Thus,the domain of the function is $x \in [1, 4]$.

(D) The function is defined as $f(x) = \cos^{-1}[x]$.
We know that the domain of $\cos^{-1}(u)$ is $u \in [-1, 1]$.
Therefore,for $\cos^{-1}[x]$ to be defined,we must have $-1 \leq [x] \leq 1$.
Since $[x]$ is the greatest integer function,the possible integer values for $[x]$ are $-1, 0, 1$.
If $[x] = -1$,then $-1 \leq x < 0$.
If $[x] = 0$,then $0 \leq x < 1$.
If $[x] = 1$,then $1 \leq x < 2$.
Combining these intervals,we get $-1 \leq x < 2$.
Thus,the domain is $x \in [-1, 2)$.

(A) We have the function $f(x) = \cos^{-1} \sqrt{x-1}$.
For the function $\cos^{-1}(u)$ to be defined,the argument $u$ must satisfy $u \in [-1, 1]$.
Since $\sqrt{x-1}$ is always non-negative,the condition becomes $0 \leq \sqrt{x-1} \leq 1$.
Squaring the inequality,we get $0 \leq x-1 \leq 1$.
Adding $1$ to all parts,we obtain $1 \leq x \leq 2$.
Therefore,the domain of $f(x)$ is $[1, 2]$.

(C) For $A$,the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$. Since $\sqrt{x^2+x+1} \ge 0$,we require $0 \le x^2+x+1 \le 1$.
Solving $x^2+x+1 \ge 0$: The discriminant $D = 1-4 = -3 < 0$,so $x^2+x+1 > 0$ for all $x \in R$.
Solving $x^2+x+1 \le 1$: $x^2+x \le 0 \implies x(x+1) \le 0$,so $x \in [-1, 0]$. Thus,$A = [-1, 0]$.
For $B$,we find the range of $y = \sin^{-1}(\sqrt{x^2+x+1})$ for $x \in [-1, 0]$.
Let $f(x) = x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4}$.
For $x \in [-1, 0]$,the minimum value of $f(x)$ is $\frac{3}{4}$ (at $x = -1/2$) and the maximum value is $1$ (at $x = -1$ or $x = 0$).
Thus,$\sqrt{x^2+x+1} \in [\sqrt{3}/2, 1]$.
Then $y = \sin^{-1}(\sqrt{x^2+x+1}) \in [\sin^{-1}(\sqrt{3}/2), \sin^{-1}(1)] = [\frac{\pi}{3}, \frac{\pi}{2}]$.
So $B = [\frac{\pi}{3}, \frac{\pi}{2}]$.
Checking options: $A = [-1, 0]$ and $B = [\frac{\pi}{3}, \frac{\pi}{2}]$.
$A \cap B = \phi$ (empty set).
$A \cap B^C = A \setminus B = [-1, 0] \setminus [\frac{\pi}{3}, \frac{\pi}{2}] = [-1, 0]$.
$A^C \cap B = B \setminus A = [\frac{\pi}{3}, \frac{\pi}{2}] \setminus [-1, 0] = [\frac{\pi}{3}, \frac{\pi}{2}]$.
Therefore,option $C$ is correct.

(B) Let $g(x) = 9x^2 - 12x + 22$.
We can rewrite this as $g(x) = (3x - 2)^2 + 18$.
The minimum value of $g(x)$ is $18$ when $3x - 2 = 0$,i.e.,$x = \frac{2}{3}$.
As $x \to \pm \infty$,$g(x) \to \infty$.
Thus,the range of $g(x)$ is $[18, \infty)$.
Consequently,the range of $\sqrt{g(x)}$ is $[\sqrt{18}, \infty) = [3\sqrt{2}, \infty)$.
Now,consider the argument of $\operatorname{Cos}^{-1}$,which is $u = \frac{3}{\sqrt{g(x)}}$.
Since $\sqrt{g(x)} \geq 3\sqrt{2}$,we have $0 < \frac{3}{\sqrt{g(x)}} \leq \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.
So,$u \in (0, \frac{1}{\sqrt{2}}]$.
Since $\operatorname{Cos}^{-1}(u)$ is a decreasing function,the range of $f(x) = \operatorname{Cos}^{-1}(u)$ is $[\operatorname{Cos}^{-1}(\frac{1}{\sqrt{2}}), \operatorname{Cos}^{-1}(0))$.
This simplifies to $[\frac{\pi}{4}, \frac{\pi}{2})$.
Therefore,the correct option is $B$.

(B) For the function $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,both terms must be defined.
For $\sqrt{2x - 1}$ to be defined,we must have $2x - 1 \geq 0$,which implies $x \geq \frac{1}{2}$.
For $\cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{2x - 1}{3} \leq 1$.
Multiplying by $3$,we get $-3 \leq 2x - 1 \leq 3$.
Adding $1$ to all parts,we get $-2 \leq 2x \leq 4$.
Dividing by $2$,we get $-1 \leq x \leq 2$.
The domain is the intersection of $x \geq \frac{1}{2}$ and $-1 \leq x \leq 2$.
Thus,the domain is $\left[\frac{1}{2}, 2\right]$.

(A) Given $f(x) = \sin^{-1} ( \frac{1 + x^2}{2 x} ) + \cos^{-1} ( \frac{2 x}{1 + x^2} )$.
For $\sin^{-1} ( \frac{1 + x^2}{2 x} )$ to be defined,we must have $| \frac{1 + x^2}{2 x} | \leq 1$.
Since $1 + x^2 \geq 2|x|$ for all $x \in R$,the condition $| \frac{1 + x^2}{2 x} | \leq 1$ holds only when $|x| = 1$,i.e.,$x = 1$ or $x = -1$.
If $x = 1$,$f(1) = \sin^{-1}(1) + \cos^{-1}(1) = \frac{\pi}{2} + 0 = \frac{\pi}{2}$.
If $x = -1$,$f(-1) = \sin^{-1}(-1) + \cos^{-1}(-1) = -\frac{\pi}{2} + \pi = \frac{\pi}{2}$.
Thus,the domain is $\{ -1, 1 \}$ and the range is $\{ \frac{\pi}{2} \}$.

(C) We know that for $x \in [-1, 1]$,$\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$.
Given expression is $f(x) = \operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x + \operatorname{Tan}^{-1} x$.
Substituting the identity,we get $f(x) = \frac{\pi}{2} + \operatorname{Tan}^{-1} x$.
The domain of $\operatorname{Sin}^{-1} x$ and $\operatorname{Cos}^{-1} x$ is $[-1, 1]$.
Therefore,the domain of $f(x)$ is $[-1, 1]$.
We calculate the values at the boundaries:
For $x = -1$,$f(-1) = \frac{\pi}{2} + \operatorname{Tan}^{-1}(-1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
For $x = 1$,$f(1) = \frac{\pi}{2} + \operatorname{Tan}^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
Since $\operatorname{Tan}^{-1} x$ is a strictly increasing function,the range of $f(x)$ is $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$.

(C) The derivative of $y = \operatorname{cosec}^{-1}(x)$ is given by $\frac{dy}{dx} = \frac{-1}{|x| \sqrt{x^2-1}}$.
For the derivative to be defined,the expression inside the square root must be strictly positive,i.e.,$x^2 - 1 > 0$,which implies $|x| > 1$.
However,the range of $y = \operatorname{cosec}^{-1}(x)$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$.
Since the derivative $\frac{dy}{dx}$ is defined for all $x$ in the domain of the function excluding the endpoints where the derivative might not exist or is undefined,we consider the open interval for $y$.
The values of $y$ corresponding to the domain of the derivative are $y \in \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right)$.

(C) Let $g(x) = x^2+x+1$. We can write this as $g(x) = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}$.
Since $\left(x+\frac{1}{2}\right)^2 \ge 0$,the minimum value of $g(x)$ is $\frac{3}{4}$.
Thus,the range of $\sqrt{g(x)}$ is $\left[\sqrt{\frac{3}{4}}, \infty\right) = \left[\frac{\sqrt{3}}{2}, \infty\right)$.
However,the domain of $\operatorname{Sin}^{-1}(u)$ is $u \in [-1, 1]$.
Therefore,we must have $\frac{\sqrt{3}}{2} \le \sqrt{x^2+x+1} \le 1$.
Squaring the inequality,we get $\frac{3}{4} \le x^2+x+1 \le 1$.
Since the function $f(u) = \operatorname{Sin}^{-1}(u)$ is an increasing function,the range of $f(x)$ is $\left[\operatorname{Sin}^{-1}\left(\frac{\sqrt{3}}{2}\right), \operatorname{Sin}^{-1}(1)\right]$.
This simplifies to $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$.