Question

IF $a\in R$ and the equation $(a-2)(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$
has no integrate solution and has exactly one solution in (2, 3), then a lies in

• A. (-1, 2)
• B. (0, 1)
• C. (-1, 0)
• D. (2, 3)

Answer 1

The correct option is C (-1, 0)

Given equation,

$(a-2){(x-\left[x\right])}^2+2(x-\left[x\right])+a^2=0$

Given equation has no integer solution

and $aϵR$ & $(2,3)$ exactly one solution $(2,3)$

If $xϵ(2,3)$ then $x-\left[x\right]ϵ(0,1)$

i.e., if one value of x lies in $(2,3)$. Then, one value of $x-\left[x\right]$ lies in $(0,1)$

Let $x-\left[x\right]=y$ lies in $(0,1)$

$\therefore f\left(y\right)=(a-2)y^2+2y+a^2=0$ has one solution of y in $(0,1)$

$\therefore f\left(0\right)$ & $f\left(1\right)$ has opposite sign

$f\left(0\right)=a^2>0$

$\therefore f\left(1\right)<0$

$(a-2)+2+a^2<0$

$a^2+a<0$

$\therefore aϵ(-1,0)$