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For all real values of $\mathrm{x}$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is
(a) 0
(b) 1
(c) 3
(d) $\frac{1}{3}$
For all real values of $\mathrm{x}$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is
(a) 0
(b) 1
(c) 3
(d) $\frac{1}{3}$
Solution:
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Verified Answer
(d) Let $\mathrm{y}=\frac{1-\mathrm{x}+\mathrm{x}^2}{1+\mathrm{x}+\mathrm{x}^2}$
$\frac{d y}{d x}=\frac{(-1+2 x)\left(1+x+x^2\right)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)}$
Numberator of $\frac{d y}{d x}=2(x-1)(x+1)$, $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2(\mathrm{x}-1)(\mathrm{x}+1)}{\left(\mathrm{x}^2+\mathrm{x}+1\right)^2} ; \quad \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad$ at $\mathrm{x}=1,-1$
At $x=1, \frac{d y}{d x}$ changes sign from $-$ ve to $+v e$
$\therefore \quad y$ is minimum at $x=1$
Minimum value of $\frac{1-x+x^2}{1+x+x^2}=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
$\frac{d y}{d x}=\frac{(-1+2 x)\left(1+x+x^2\right)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)}$
Numberator of $\frac{d y}{d x}=2(x-1)(x+1)$, $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2(\mathrm{x}-1)(\mathrm{x}+1)}{\left(\mathrm{x}^2+\mathrm{x}+1\right)^2} ; \quad \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad$ at $\mathrm{x}=1,-1$
At $x=1, \frac{d y}{d x}$ changes sign from $-$ ve to $+v e$
$\therefore \quad y$ is minimum at $x=1$
Minimum value of $\frac{1-x+x^2}{1+x+x^2}=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
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