Match the statements given in Column-I with the intervals/union of intervals given in Column-II. Column-I Column-II (A) The set { is a complex number, } is (p) (B) The domain of the function is (q) (C) If then the set is (r) (D) If then

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Solution

Q.
Match the statements given in Column-I with the intervals/union of intervals given in Column-II. [2011]

Column-I Column-II

(A) The set { $Re (\frac{2 i z}{1 - z^{2}}) :$ $z$ is a complex number,
$| z | = 1,$ $z \neq \pm 1$ } is (p) $(- \infty, - 1) \cup (1, \infty)$

(B) The domain of the function
$f (x) = \sin^{- 1} (\frac{8 {(3)}^{x - 2}}{1 - 3^{2 (x - 1)}})$ is (q) $(- \infty, - 0) \cup (0, \infty)$

(C) If
$f (θ) =$ $| \begin{matrix} 1 & \tan θ & 1 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} |$ $,$
then the set
${f (θ) : 0 \leq θ < \frac{π}{2}}$ is (r) $[2, \infty)$

(D) If $f (x) =$ $x^{3 / 2}$
$(3 x - 10),$ $x \geq 0$ then
$f (x)$ is increasing in (s) $(- \infty, - 1] \cup [1, \infty)$

(t) $(- \infty, 0] \cup [2, \infty)$

1 (A) → (s), (B) → (t), (C) → (r), (D) → (r)

2 (A) → (r), (B) → (t), (C) → (r), (D) → (s)

3 (A) → (r), (B) → (r), (C) → (t), (D) → (s)

4 (A) → (t), (B) → (r), (C) → (r), (D) → (s)

Ans.
(1)

(A) → (s), (B) → (t), (C) → (r), (D) → (r)

$(A)  Let z = x + i y . Given that | z | = 1, i.e. x^{2} + y^{2} = 1, x \neq \pm 1$

Then   $Re (\frac{2 i z}{1 - z^{2}})$ $= Re (\frac{2 i z}{z . \bar{z} - z^{2}})$

$= Re (\frac{2 i}{\bar{z} - z})$ $= Re (\frac{2 i}{- 2 i y})$ $= Re (\frac{- 1}{y})$ $= - \frac{1}{y},$

where, $x = \sqrt{1 - y^{2}}$

$- 1 \leq y \leq 1 \Rightarrow - \frac{1}{y} \geq 1 or - \frac{1}{y} \leq - 1$

$∴$ $Re (\frac{2 i z}{1 - z^{2}}) \in (- \infty, - 1] \cup [1, \infty)$

$∴$    $A \to s$

$(B)  For the domain of f (x) = \sin^{- 1} (\frac{8 \cdot 3^{x - 2}}{1 - 3^{2 (x - 1)}})$

We should have

$- 1 \leq (\frac{8 \cdot 3^{x - 2}}{1 - 3^{2 (x - 1)}}) \leq 1$ $\Rightarrow - 1 \leq \frac{8 \cdot 3^{x}}{9 - 3^{2 x}} \leq 1$

$\Rightarrow \frac{8 \cdot 3^{x}}{9 - 3^{2 x}} \geq - 1$ $\Rightarrow \frac{8 \cdot 3^{x} + 9 - 3^{2 x}}{9 - 3^{2 x}} \geq 0$

$\Rightarrow \frac{(3^{x} - 9) (3^{x} + 1)}{(3^{x} - 3) (3^{x} + 3)} \geq 0$

We know that $3^{x} > 0$

Hence, $x \in (- \infty, 0) \cup (2, \infty)$    $\dots (i)$

Also, $\frac{8 \cdot 3^{x}}{9 - 3^{2 x}} \leq 1$ $\Rightarrow \frac{8 \cdot 3^{x} - 9 + 3^{2 x}}{9 - 3^{2 x}} \leq 0$

$\Rightarrow \frac{(3^{x} + 9) (3^{x} - 1)}{(3^{x} - 3) (3^{x} + 3)} \geq 0$

Hence, $x \in (- \infty, 0] \cup (1, \infty)$    $\dots (i i)$

From (i) and (ii), we get $x \in (- \infty, 0] \cup [2, \infty)$

$∴$    $B \to t$

$(C)$ $f (θ) =$ $| \begin{matrix} 1 & \tan θ & 1 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} |$

Applying   $R_{1} \to R_{1} + R_{3}$ $=$ $| \begin{matrix} 0 & 0 & 2 \\ - \tan θ & 1 & \tan θ \\ - 1 & - \tan θ & 1 \end{matrix} |$

$= 2 (1 + \tan^{2} θ) = 2 \sec^{2} θ \geq 2,$    $0 \leq θ < \frac{π}{2}$

$∴$ $C \to r$

$(D) f (x) = x^{3 / 2} (3 x - 10),$    $x \geq 0$

$∴$ $f' (x) = \frac{3}{2} x^{1 / 2} (3 x - 10) + x^{3 / 2} (3)$

For $f (x)$ to be increasing, $f' (x) \geq 0$

$\Rightarrow 3 x^{1 / 2} [\frac{3 x - 10}{2} + x] \geq 0$ $\Rightarrow 3 x^{1 / 2} (5 x - 10) \geq 0$

Hence, $f (x) is increasing on [2, \infty)$

$∴$    $D \to r$

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