TS EAMCET · Maths · Quadratic Equation
For all real values of \(\mathrm{x}\), the minimum value of \(\frac{1-x+x^2}{1+x+x^2}\) is
- A \(0\)
- B \(\frac{1}{3}\)
- C \(1\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & f=\frac{1-x+x^2}{1+x+x^2}=\frac{x^2+x+1-2 x}{x^2+x+1} \\ & f=1-\frac{2 x}{x^2+x+1} \end{aligned} \] For \(f\) to be minimum, \(\frac{2 x}{x^2+x+1}\) must be maximum…
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