TS EAMCET · Maths · Functions
\(f\) is real valued function satisfying the relation \(f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}\). If \(f\left(x+\frac{1}{x}\right)=1\) then \(x=\)
- A \(\pm 2\)
- B \(\pm 1\)
- C \(\pm 3\)
- D \(\pm 6\)
Answer & Solution
Correct Answer
(B) \(\pm 1\)
Step-by-step Solution
Detailed explanation
Given that \(f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}\) \(f\left(3 x+\frac{1}{2 x}\right)=9 x^2+\frac{1}{4 x^2}+2 \cdot 3 x \cdot \frac{1}{2 x}-3\) \(\Rightarrow f\left(3 x+\frac{1}{2 x}\right)=\left(3 x+\frac{1}{2 x}\right)^2-3\) \(\therefore f(x)=x^2-3\)…
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