Sets, Relations & Functions
Solving Logarithmic / Radical Equations
nta_pyq_2025_apr
Grade 11
Let $A = \left\{x \in (0,\pi) - \left\{\frac{\pi}{2}\right\} : \log_{2/\pi}|\sin x| + \log_{2/\pi}|\cos x| = 2\right\}$ and $B = \{x \geq 0 : \sqrt{x}(\sqrt{x}-4) - 3|\sqrt{x}-2| + 6 = 0\}$. Then $n(A \cup B)$ is equal to:
Step-by-Step Solution
Key Concept: For A: combine logs to get $|\sin 2x| = 8/\pi^2$, which has 4 solutions in $(0,\pi)-\{\pi/2\}$. For B: split into cases $\sqrt{x}<2$ and $\sqrt{x}\geq2$, giving $x\in\{0,1,9,16\}$.
A: $\log_{2/\pi}|\sin x \cos x| = 2 \Rightarrow |\sin 2x| = \frac{8}{\pi^2}$ — 4 solutions in $(0,\pi)\setminus\{\pi/2\}$. B: Let $t=\sqrt{x}$: for $t<2$: $t^2-t=0\Rightarrow t=0,1$; for $t\geq2$: $t^2-7t+12=0\Rightarrow t=3,4$. So $x=0,1,9,16$ — 4 solutions. $n(A\cup B) = 4+4 = 8$.
Correct Answer: 8