Let A = { x ( 0 , ) – { 2 } : log ( 2 / ) | sin x | + log ( 2 / ) | cos x | = 2 } and B = { x 0 : x ( x – 4 ) – 3 | x – 2 | + 6 = 0 } . Then n ( A B ) is equal to : [2025]

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Solution

Q.
Let $A = {x \in (0, π) - {\frac{π}{2}} : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2}$ and $B = {x \geq 0 : \sqrt{x} (\sqrt{x} - 4) - 3 | \sqrt{x} - 2 | + 6 = 0}$ .

Then $n (A \cup B)$ is equal to : [2025]

1 4

2 2

3 8

4 6

Ans.
(3)

For $A : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2$

$\Rightarrow \log_{(2 / π)} (| \sin x \cdot \cos x |) = 2$

$\Rightarrow | \sin x \cdot \cos x | = {(\frac{2}{π})}^{2}$

$\Rightarrow | 2 \sin x \cdot \cos x | = \frac{8}{π^{2}} \Rightarrow | \sin 2 x | = \frac{8}{π^{2}}$

The region is given by

Since, line $y = \frac{8}{π^{2}}$ cuts the graph of $y =$ $| \sin 2 x |$ four times. Thus, there are four solutions for A.

For $B : \sqrt{x} (\sqrt{x} - 4) - 3 | \sqrt{x} - 2 | + 6 = 0$

When $x < 4, \sqrt{x} (\sqrt{x} - 4) + 3 (\sqrt{x} - 2) + 6 = 0$

$\Rightarrow x - 4 \sqrt{x} + 3 \sqrt{x} - 6 + 6 = 0 \Rightarrow x - \sqrt{x} = 0$

$\Rightarrow \sqrt{x} (\sqrt{x} - 1) = 0 \Rightarrow \sqrt{x} = 0 or 1$

$\Rightarrow x = 0 or 1$

When $x > 4, \sqrt{x} (\sqrt{x} - 4) - 3 (\sqrt{x} - 2) + 6 = 0$

$\Rightarrow x - 4 \sqrt{x} - 3 \sqrt{x} + 6 + 6 = 0 \Rightarrow x - 7 \sqrt{x} + 12 = 0$

$\Rightarrow (\sqrt{x} - 3) (\sqrt{x} - 4) = 0 \Rightarrow \sqrt{x} = 3 or 4$

$\Rightarrow x = 9 or 16$

Thus, there are four solutions for B as well.

Now, $n (A \cup B) = n (A) + n (B) = 4 + 4 = 8$

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