Let A= \ x in(0, π)- \ π/2 \ : log (2 / π) sin x +log (2 / π) cos x =2 \ and B=\ x geq 0: √(x)(√(x)-4)-3 √(x)-2 +6=0\ . Then n ( A cup B ) is equal to:

Mathematics · Basic Maths

JEE Main 2025 — 24 January, Evening Shift — Question 5

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 $\mathrm{n}(\mathrm{A} \cup \mathrm{B})$ is equal to:

A : $\log _{2/ \pi}|\sin x|+\log _{2/ \pi}|\cos x|=2$

$\Rightarrow \log _{2/ \pi}(|\sin x . \cos x|)=2$

$\Rightarrow|\sin 2 \mathrm{x}|=\frac{8}{\pi^{2}}$

Number of solution 4

B : let $\sqrt{\mathrm{x}}=\mathrm{t}<2$

Then $\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}-4)+3(\sqrt{\mathrm{x}}-2)+6=0$

$\Rightarrow \mathrm{t}^{2}-4 \mathrm{t}+3 \mathrm{t}-6+6=0$

$\Rightarrow \mathrm{t}^{2}-\mathrm{t}=0; \mathrm{t}=0, \mathrm{t}=1$

$\mathrm{x}=0, \mathrm{x}=1$

again let $\sqrt{\mathrm{x}}=\mathrm{t}>2$

then $\mathrm{t}^{2}-4 \mathrm{t}-3 \mathrm{t}+6+6=0$

$\Rightarrow \mathrm{t}^{2}-7 \mathrm{t}+12=0$

$\Rightarrow \mathrm{t}=3,4$ $\mathrm{x}=9,16$

Total number of solutions

$\mathrm{n}(\mathrm{A} \cup \mathrm{B})=4+4=8$

Answer key and solution verified before publishing.

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