Question

The curve $y^2=(x-a)(x-b)$, $a,b>0$ and $a>b$ does not exist for:

• A. $x\ge a$
• B. $x=b$
• C. $b<x<a$
• D. $x=a$

Answer 1

The correct option is C $b<x<a$

Given curve is, $y^2=(x-a)(x-b)$

Since we know that, square of any real number is always non-negative

So LHS of given equation is always $\ge 0$

So for above equation to be well defined , RHS should also be $\ge 0$

Hence above equation does not exist if $RHS<0$

$\Rightarrow (x-a)(x-b)<0$

$\Rightarrow b<x<a$, since $a>b$