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  4. If the graph of the function y=(a - b)2x2+2(a + b - 2 c)x+1(∀ a ≠ b) is strictly above the x -axis, then

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Q. If the graph of the function $y=\left(a - b\right)^{2}x^{2}+2\left(a + b - 2 c\right)x+1\left(\forall a \neq b\right)$ is strictly above the $x$ -axis, then

NTA AbhyasNTA Abhyas 2022

Solution:

$D=B^{2}-4AC$
$=\left(2 \left(a + b - 2 c\right)\right)^{2}-4\left(a - b\right)^{2}$
$=4\left\{\left(a + b - 2 c\right)^{2} - \left(a - b\right)^{2}\right\}$
$=4\left(a + b - 2 c - a + b\right)\left(a + b - 2 c + a - b\right)$
$=4\left(2 b - 2 c\right)\left(2 a - 2 c\right)$
$=16\left(b - c\right)\left(a - c\right)$
$=16\left(c - b\right)\left(c - a\right)$
If $c$ lies between $a$ and $b,$ then $D$ is negative. Hence, the roots will be imaginary and the graph will be entirely above the $x$ -axis as the coefficient of $x^{2}$ is positive.