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The inequality $\frac{2-gx+x^{2}}{1-x+x^{2}}\leq 3$ is true for all the value of $x$ if and only if

  1. $1\leq g\leq 7$
  2. $-1\leq g\leq 1$
  3. $-6\leq g\leq 7$
  4. $-1\leq g\leq 7$
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$\frac{2 - gx + x^{2}}{1-x+x^{2}} \leqslant 3$

$2 - gx + x^{2} \leqslant 3 - 3x+ 3 x^{2}$

$2x^{2} + (g-3)x + 1 \geqslant 0$

$\text{For real roots, }$ $b^{2}-4ac \leqslant 0$

$(g-3)^{2}-4*2*1 \leqslant 0$

$g^{2}+9 - 6g - 8 \leqslant 0$

$g^{2} - 6g +1 \leqslant 0$

$g$ $lies$ $between $ $ 3 \pm 2\sqrt2$

 

No options are matching. really.

 

But if the question is $-x^{2}$ in the numerator , then g is coming in the range OPTION D
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$$\begin{align*} &\Rightarrow \frac{2-gx+x^{2}}{1-x+x^{2}}\leq 3 \\ &\Rightarrow 2-gx+x^{2} \leq 3 - 3x + 3x^2 \\ &\Rightarrow 2x^2 + \left ( g-3 \right )x + 1 \geq 0 \\ \end{align*}$$

So root will be imaginary and the curve never touches the $X$ axis. Like the following curve :

$$\begin{align*} &\Rightarrow B^2 - 4AC < 0 \\ &\Rightarrow \left ( g-3 \right )^2 - 4.2.1 <0 \\ &\Rightarrow -\sqrt{8} < \left ( g-3 \right ) < +\sqrt{8} \\ &\Rightarrow -\sqrt{8}+3 < g < +\sqrt{8}+3 \\ &\Rightarrow 0.1715 < g < +5.83 \\ &\Rightarrow \text{None of the options seem to match:}  \end{align*}$$

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