Match Complex Numbers Functions Determinants and Derivatives to Intervals

To match the statements given in Column I with the correct intervals/union of intervals in Column II, we evaluate each part step-by-step:

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Part (A)

We are given the set:
$\left\{\text{Re}\left(\frac{2iz}{1-z^2}\right) : z \text{ is a complex number, } |z| = 1, z \neq \pm 1\right\}$

Since $|z| = 1$ and $z \neq \pm 1$, we can express $z$ in polar form as $z = e^{i\theta} = \cos\theta + i\sin\theta$, where $\theta \in \mathbb{R} \setminus \{k\pi : k \in \mathbb{Z}\}$.

Substituting $z = e^{i\theta}$ into the given expression:
$w = \frac{2iz}{1-z^2} = \frac{2i e^{i\theta}}{1 - e^{2i\theta}}$

Dividing the numerator and denominator by $e^{i\theta}$:
$w = \frac{2i}{e^{-i\theta} - e^{i\theta}} = \frac{2i}{-2i\sin\theta} = -\frac{1}{\sin\theta}$

Since $\theta \neq k\pi$, the range of $\sin\theta$ is $[-1, 0) \cup (0, 1]$.
Consequently, the value of $w$ lies in:
$w \in (-\infty, -1] \cup [1, \infty)$

Since $w$ is purely real, $\text{Re}(w) = w \in (-\infty, -1] \cup [1, \infty)$.

Thus, (A) matches with (s).

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Part (B)

We are given the function:
$f(x) = \sin^{-1}\left(\frac{8(3)^{x-2}}{1 - 3^{2(x-1)}}\right)$

For $f(x)$ to be defined, the argument of $\sin^{-1}$ must satisfy:
$-1 \le \frac{8 \cdot 3^{x-2}}{1 - 3^{2(x-1)}} \le 1 \implies \left| \frac{8 \cdot 3^{x-2}}{1 - 3^{2(x-1)}} \right| \le 1$

Let $t = 3^{x-1} > 0$. Then $3^{x-2} = \frac{t}{3}$ and $3^{2(x-1)} = t^2$. Substituting these yields:
$\left| \frac{\frac{8t}{3}}{1 - t^2} \right| \le 1 \implies 8t \le 3|1 - t^2| \quad (t \neq 1)$

We consider two cases based on the value of $t$:

1. Case 1: $0 < t < 1$ (i.e., $1 - t^2 > 0$)
   $8t \le 3(1 - t^2) \implies 3t^2 + 8t - 3 \le 0 \implies (3t - 1)(t + 3) \le 0$
   Since $t > 0$, this inequality simplifies to:
   $0 < t \le \frac{1}{3} \implies 3^{x-1} \le 3^{-1} \implies x - 1 \le -1 \implies x \le 0$
   So, $x \in (-\infty, 0]$.

2. Case 2: $t > 1$ (i.e., $1 - t^2 < 0$)
   $8t \le 3(t^2 - 1) \implies 3t^2 - 8t - 3 \ge 0 \implies (3t + 1)(t - 3) \ge 0$
   Since $t > 1$, this inequality simplifies to:
   $t \ge 3 \implies 3^{x-1} \ge 3^1 \implies x - 1 \ge 1 \implies x \ge 2$
   So, $x \in [2, \infty)$.

Combining both cases, the domain of $f(x)$ is $(-\infty, 0] \cup [2, \infty)$.

Thus, (B) matches with (t).

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Part (C)

We are given the determinant:

$$f(\theta) = \begin{vmatrix} 1 & \tan\theta & 1 \\ -\tan\theta & 1 & \tan\theta \\ -1 & -\tan\theta & 1 \end{vmatrix}$$

Expanding the determinant along the first row:
$f(\theta) = 1 \cdot (1 + \tan^2\theta) - \tan\theta \cdot (-\tan\theta + \tan\theta) + 1 \cdot (\tan^2\theta + 1)$
$f(\theta) = (1 + \tan^2\theta) - 0 + (1 + \tan^2\theta) = 2(1 + \tan^2\theta) = 2\sec^2\theta$

For $0 \le \theta < \frac{\pi}{2}$, we have $\sec\theta \ge 1 \implies \sec^2\theta \in [1, \infty)$.
Therefore:
$f(\theta) = 2\sec^2\theta \in [2, \infty)$

Thus, (C) matches with (r).

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Part (D)

We are given the function:
$f(x) = x^{3/2}(3x - 10) = 3x^{5/2} - 10x^{3/2}, \quad x \ge 0$

Differentiating $f(x)$ with respect to $x$:
$f'(x) = 3 \cdot \frac{5}{2}x^{3/2} - 10 \cdot \frac{3}{2}x^{1/2} = \frac{15}{2}x^{3/2} - 15x^{1/2} = \frac{15}{2}\sqrt{x}(x - 2)$

For $f(x)$ to be increasing, we require $f'(x) \ge 0$:
$\frac{15}{2}\sqrt{x}(x - 2) \ge 0$

Since $\sqrt{x} \ge 0$ for all $x \ge 0$, $f'(x) \ge 0$ holds when:
$x - 2 \ge 0 \implies x \ge 2$

Thus, $f(x)$ is increasing in $[2, \infty)$.

Thus, (D) matches with (r).

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Final Match:

• (A) $\rightarrow$ (s)
• (B) $\rightarrow$ (t)
• (C) $\rightarrow$ (r)
• (D) $\rightarrow$ (r)