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If \((f(x))^2=f\left(x^2\right)+f(1)\) holds good, then find \(f(x)\)
Options:
Solution:
2123 Upvotes
Verified Answer
The correct answer is:
\(x+\frac{1}{x}\)
Given functional relation
\((f(x))^2=f\left(x^2\right)+f(\mathrm{l})\)
On putting \(f(x)=x+\frac{1}{x}\), the RHS is
\(x^2+\frac{1}{x^2}+2=\left(x+\frac{1}{x}\right)^2=(f(x))^2=\text { LHS. }\)
Hence, option (a) is correct.
\((f(x))^2=f\left(x^2\right)+f(\mathrm{l})\)
On putting \(f(x)=x+\frac{1}{x}\), the RHS is
\(x^2+\frac{1}{x^2}+2=\left(x+\frac{1}{x}\right)^2=(f(x))^2=\text { LHS. }\)
Hence, option (a) is correct.
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