AP EAMCET · Maths · Quadratic Equation
\(f(x)\) is a quadratic polynomial satisfying the condition \(f(x)+f\left(\frac{1}{x}\right)=f(x) f\left(\frac{1}{x}\right)\). If \(f(-1)=0\), then the range of \(f\) is
- A \([1, \infty)\)
- B \([-1,1]\)
- C \((-\infty, 1]\)
- D \(\mathbb{R}\)
Answer & Solution
Correct Answer
(C) \((-\infty, 1]\)
Step-by-step Solution
Detailed explanation
\((f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=1\) Let \(f(x)=ax^2+bx+c\) for \(a \ne 0\). \((ax^2+bx+c-1)\left(a\frac{1}{x^2}+b\frac{1}{x}+c-1\right)=1\) \((ax^2+bx+c-1)\left(\frac{(c-1)x^2+bx+a}{x^2}\right)=1\) \((ax^2+bx+c-1)((c-1)x^2+bx+a) = x^2\) For this identity to…
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