Question ID: #392
Let $A=\{x\in(0,\pi)-\{\frac{\pi}{2}\}: \log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\}$ and $B=\{x\ge0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}$. Then $n(A\cup B)$ is equal to:
- (1) 4
- (2) 2
- (3) 8
- (4) 6
Solution:
First, we solve for the number of elements in set A. The given logarithmic equation simplifies as follows:
$$ \log_{2/\pi}(|\sin x||\cos x|) = 2 $$
$$ |\sin x \cos x| = \left(\frac{2}{\pi}\right)^2 = \frac{4}{\pi^2} $$
Multiplying both sides by 2, we get:
$$ |2\sin x \cos x| = \frac{8}{\pi^2} \Rightarrow |\sin 2x| = \frac{8}{\pi^2} $$
Thus, $n(A) = 4$.
Next, we solve for set B. Let $\sqrt{x} = t$ where $t \ge 0$. The equation becomes:
$$ t(t-4) – 3|t-2| + 6 = 0 \Rightarrow t^2 – 4t – 3|t-2| + 6 = 0 $$
If $0 \le t < 2$, then $|t-2| = 2-t$. The equation is: $$ t^2 - 4t - 3(2-t) + 6 = 0 \Rightarrow t^2 - t = 0 \Rightarrow t = 0, 1 $$ If $t \ge 2$, then $|t-2| = t-2$. The equation is: $$ t^2 - 4t - 3(t-2) + 6 = 0 \Rightarrow t^2 - 7t + 12 = 0 \Rightarrow t = 3, 4 $$ The valid values for $t$ are $\{0, 1, 3, 4\}$, which give 4 distinct values for $x$. Thus, $n(B) = 4$.
Since set A consists of trigonometric solutions and set B consists of algebraic solutions (integers), they are disjoint.
$$ n(A \cup B) = n(A) + n(B) = 4 + 4 = 8 $$
Ans. (3)
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