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If the graph of the function $y=(a-b)^2 x^2+2(a+b-2 c) x+1(\forall a \neq b)$ is strictly above the $x$-axis, then
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The correct answer is:
$\mathrm{a} \lt \mathrm{c} \lt \mathrm{b}$
$\mathrm{D}=\mathrm{B}^2-4 \mathrm{AC}$
$=(2(a+b-2 c))^2-4(a-b)^2$
$=4\left\{(\mathrm{a}+\mathrm{b}-2 \mathrm{c})^2-(\mathrm{a}-\mathrm{b})^2\right\}$
$=4(a+b-2 c-a+b)(a+b-2 c+a-b)$
$=4(2 b-2 c)(2 a-2 c)$
$=16(b-c)(a-c)$
$=16(c-b)(c-a)$
If $\mathrm{c}$ lies between $\mathrm{a}$ and $\mathrm{b}$, then $\mathrm{D}$ is negative. Hence, the roots will be imaginary and the graph will be entirely above the $\mathrm{x}$-axis as the coefficient of $x^2$ is positive.
$=(2(a+b-2 c))^2-4(a-b)^2$
$=4\left\{(\mathrm{a}+\mathrm{b}-2 \mathrm{c})^2-(\mathrm{a}-\mathrm{b})^2\right\}$
$=4(a+b-2 c-a+b)(a+b-2 c+a-b)$
$=4(2 b-2 c)(2 a-2 c)$
$=16(b-c)(a-c)$
$=16(c-b)(c-a)$
If $\mathrm{c}$ lies between $\mathrm{a}$ and $\mathrm{b}$, then $\mathrm{D}$ is negative. Hence, the roots will be imaginary and the graph will be entirely above the $\mathrm{x}$-axis as the coefficient of $x^2$ is positive.
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