Match the statements given in Column I with the intervals/union of intervals given in Column II. Column-I Column-II (A) The set \left{\right. Re \left(\right. \frac{2 iz}{1 - z^{2}} \left.\right) : \textrm{ }\textrm{ } z \textrm{ }\textrm{ } is \textrm{ }\textrm{ } a \textrm{ }\textrm{ } complex \textrm{ }\textrm{ } number , \textrm{ }\textrm{ } \left|\right. z \left|\right. = 1 , \textrm{ }\textrm{ } z \neq \textrm{ } \pm 1 \left.\right} (P) \left(\right. - \infty , - 1 \left.\right) \cup \left(\right. 1 , \infty \left.\right) (B) The domain of the function f \left(\right. x \left.\right) = \left(sin\right)^{- 1} \left(\right. \frac{8 \left(\left(\right. 3 \left.\right)\right)^{x - 2}}{1 - 3^{2 \left(\right. x - 1 \left.\right)}} \left.\right) (Q) \left(\right. - \infty , 0 \left.\right) \cup \left(\right. 0 , \infty \left.\right) (C) If \textrm{ }\textrm{ } f \left(\right. \theta \left.\right) = \left|\right. \textrm{ } 1 & tanθ & 1 \\ - tanθ & 1 & tanθ \\ - 1 & - tanθ & 1 \textrm{ } \left|\right. , then the set \left{\right. f \left(\right. \theta \left.\right) : \textrm{ }\textrm{ }\textrm{ } 0 \leq \textrm{ }\textrm{ } \theta \textrm{ }\textrm{ } < \textrm{ }\textrm{ } \frac{\pi}{2} \left.\right} , is (R) [2, \infty) (D) If \textrm{ }\textrm{ } f \left(\right. x \left.\right) = x^{\frac{3}{2}} \left(\right. 3 x - 10 \left.\right) , x \geq 0 then f (x) is increasing in (S) \left(\right. - \infty , 0 \left]\right. \cup \left[\right. 2 , \infty \left.\right) (T) \left(\right. - \infty , 0 \left]\right. \cup \left[\right. 2 , \infty \left.\right)

Engineering

Mathematics

Conjugate Modulus and Argument of a Complex Number

Question

Match the statements given in Column I with the intervals/union of intervals given in Column II.

Column-I	Column-II
(A) The set ${Re (\frac{2 iz}{1 - z^{2}}) : z is a complex number, \| z \| = 1, z \neq \pm 1}$	(P) $(- \infty, - 1) \cup (1, \infty)$
(B) The domain of the function $f (x) = \sin^{- 1} (\frac{8 {(3)}^{x - 2}}{1 - 3^{2 (x - 1)}})$	(Q) $(- \infty, 0) \cup (0, \infty)$
(C) $If f (θ) = \| \begin{matrix} 1 & tanθ & 1 \\ - tanθ & 1 & tanθ \\ - 1 & - tanθ & 1 \end{matrix} \|,$ then the set ${f (θ) : 0 \leq θ < \frac{π}{2}}, is$	(R) [2, $\infty$ )
(D) $If f (x) = x^{\frac{3}{2}} (3 x - 10), x \geq 0$ then f (x) is increasing in	(S) $(- \infty, 0] \cup [2, \infty)$
	(T) $(- \infty, 0] \cup [2, \infty)$

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(A) Let z = cos $θ$ + I sin $θ$

$so \frac{2 iz}{1 - z} = \frac{2 i (\cos + isinθ)}{1 - \cos 2 θ - isin 2 θ} = - \cos ecθ \forall θ \neq (2 n + 1) \frac{π}{2}$

$so Re (\frac{2 iz}{1 - z^{2}}) = - \cos ecθ \in (- \infty, - 1) \cup [1, \infty]$

(B) $\frac{8 \times 3^{x - 2}}{1 - 3^{2 x - 2}} = \frac{8 \times 3^{x}}{9 - 3^{2 x}}$ Let 3^x = t So f(x) = sin^–1 $(\frac{8 \times 3^{x}}{9 - 3^{2 x}}) = \sin^{- 1} (\frac{8 t}{9 - t^{2}})$

$- 1 \leq \frac{8 t}{9 - t^{2}} \leq 1 on solving$ $x \in (- \infty, 0) \cup [2, \infty] \cup [1]$

(D) f(x) = 3x5^/2 – 10x3^/2

$f' (x) = \frac{15 \sqrt{x}}{2} (x - 2)$

So f(x) is increasing for f’(x) $\leq$ 0

x $\in$ [2, $\infty$ ]

* The most appropriate answer to this question is

A $\to$ q ; B $\to$ p ; C $\to$ s ; D $\to$ s

But because of ambiguity in language, IIT has declared

A $\to$ s ; B $\to$ t ; C $\to$ r ; D $\to$ r as correct answer.