Q54 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
For all real values of $\mathrm{x}$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is
- A$0$
- Correct answerB$\frac{1}{3}$
- C$1$
- D$3$
Solution
$$
\begin{aligned}
& f=\frac{1-x+x^2}{1+x+x^2}=\frac{x^2+x+1-2 x}{x^2+x+1} \\
& f=1-\frac{2 x}{x^2+x+1}
\end{aligned}
$$
For $f$ to be minimum, $\frac{2 x}{x^2+x+1}$ must be maximum
$$
\begin{aligned}
& \frac{2 x}{x^2+x+1}=\frac{2}{\left(x+\frac{1}{x}\right)+1} \\
& \because x+\frac{1}{x} \geq 2 \\
& \therefore \quad \max \cdot \frac{2 x}{x^2+x+1}=\frac{2}{3} \\
& \therefore \quad f_{\min }=1-\frac{2}{3}=\frac{1}{3} .
\end{aligned}
$$
Hence, the correct answer is option: 'B'.