JEE Advanced · Mathematics · 2. Quadratic Equations
The value of
\(6+\log _{\frac{3}{2}}\)\(\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}} \cdots}}\right)\) is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
Let \(x=\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}}} \ldots \ldots\)
\(\Rightarrow x=\frac{1}{3 \sqrt{2}} \sqrt{4-x}\)
\(\Rightarrow 3 \sqrt{2} x=\sqrt{4-x}\)
\(\Rightarrow 18 x^{2}=4-x \)
\(\Rightarrow 18 x^{2}+x-4=0 \)
\(\Rightarrow (9 x-4)(2 x+1)=0 \)
\(\Rightarrow x=\frac{4}{9} \text { or } x=-\frac{1}{2}\)
(Not possible because log is not define for -ve value)
\(\therefore 6+\log _{\frac{3}{2}}\left(\frac{4}{9}\right)=6+\log _{\frac{3}{2}}\left(\frac{3}{2}\right)^{-2}\)
\(=6-2=4\)
\(\Rightarrow x=\frac{1}{3 \sqrt{2}} \sqrt{4-x}\)
\(\Rightarrow 3 \sqrt{2} x=\sqrt{4-x}\)
\(\Rightarrow 18 x^{2}=4-x \)
\(\Rightarrow 18 x^{2}+x-4=0 \)
\(\Rightarrow (9 x-4)(2 x+1)=0 \)
\(\Rightarrow x=\frac{4}{9} \text { or } x=-\frac{1}{2}\)
(Not possible because log is not define for -ve value)
\(\therefore 6+\log _{\frac{3}{2}}\left(\frac{4}{9}\right)=6+\log _{\frac{3}{2}}\left(\frac{3}{2}\right)^{-2}\)
\(=6-2=4\)
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