Let A = { x ∈ ( 0 , π ) − { π 2 } : log ( 2 / π ) | sin x | + log ( 2 / π ) | cos x | = 2 } and .......

Question

Let A={x∈(0,π)−{π2}:log(2/π)|sinx|+log(2/π)|cosx|=2}A={x∈(0,π)−{π2}:log(2/π)⁡|sin⁡x|+log(2/π)⁡|cos⁡x|=2} and 

B={x⩾0:√x(√x−4)−3|√x−2|+6=0}B={x⩾0:x(x−4)−3|x−2|+6=0}. 

Then n(A∪B)n(A∪B) is equal to ?

A.4
B.8  

C.6

D.2

Answer 1

 Answer: (3) 8 

\( A : \log _{2 \pi}|\sin x|+\log _{2 \pi}|\cos x|=2\)

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

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

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

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