Question

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

• A. $a<b<c$
• B. $a<c<b$
• C. $b<a<c$
• D. $c<b<a$

Answer 1

Answer: (B)

$D=B^2-4AC$
$={\left(2\left(a+b-2c\right)\right)}^2-4{\left(a-b\right)}^2$
$=4\left\{{\left(a+b-2c\right)}^2-{\left(a-b\right)}^2\right\}$
$=4\left(a+b-2c-a+b\right)\left(a+b-2c+a-b\right)$
$=4\left(2b-2c\right)\left(2a-2c\right)$
$=16\left(b-c\right)\left(a-c\right)$
$=16\left(c-b\right)\left(c-a\right)$
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.