Question

If $a \in R$ and the equation $- 3{(x - \left[ x \right])^2} + 2(x - \left[ x \right]) + {a^2} = 0$ (where $\left[ x \right]$ denotes the greatest integer $\leqslant x$ ) has no integral solution but has real solution, then all possible values of a lie in the interval:
A. $( - 1,0) \cup (0,1)$
B. $(1,2)$
C. $( - 2, - 1)$
D. $( - \infty , - 2) \cup (2,\infty )$

Answer 1

To solve this first we will convert the above equation into a simple quadratic equation, and will get its root by using quadratic formula. Putting the root in interval of 0 and 1 as it is a fractional part we will get the interval of possible values of a.

Complete step-by-step answer:
The Greatest Integer Function is denoted by $y = [x]$. For all real numbers, x, the greatest integer function returns the largest integer. less than or equal to x. In essence, it rounds down a real number to the nearest integer.
Let x - \left[ x \right] = t = \left\\{ X \right\\} , which is a fractional part function.
It lies between 0 and1, 0 \leqslant \left\\{ X \right\\} < 1
According to question,$a \in R$ and the given equation is,
$- 3{(x - \left[ x \right])^2} + 2(x - \left[ x \right]) + {a^2} = 0$
Putting $x - \left[ x \right] = t$ in the above equation we get,
$- 3{t^2} + 2t + {a^2} = 0$
Comparing the above equation with $a^2x+bx+c=0$ we get $a=-3$ , $b=2$ and $c=a^2$.
Putting the values in quadratic formula $t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ in the above equation,
$t = \dfrac{{ - 2 \pm \sqrt {{{( - 2)}^2} - 4( - 3){a^2}} }}{{2( - 3)}}$
Expanding it we get,
$t = \dfrac{{ - 2 \pm \sqrt {4 + 12{a^2}} }}{{ - 6}}$
Taking $\sqrt 4$ common from the square root term of the right hand side we get two type of solution of +2 and -2, i.e.
$t = \dfrac{{ - 2 \pm ( - 2)\sqrt {1 + 3{a^2}} }}{{ - 6}}$ and $t = \dfrac{{ - 2 \pm 2\sqrt {1 + 3{a^2}} }}{{ - 6}}$
Taking -2 common from the numerator of right hand side in both the cases we get, $t = ( - 2)\dfrac{{1 \pm \sqrt {1 + 3{a^2}} }}{{ - 6}}$ and $t = ( - 2)\dfrac{{1 \mp \sqrt {1 + 3{a^2}} }}{{ - 6}}$
Cancelling -2 with -6 of denominator we get,
$t = \dfrac{{1 \pm \sqrt {1 + 3{a^2}} }}{3}$ and $t = \dfrac{{1 \mp \sqrt {1 + 3{a^2}} }}{3}$
$\because$ x - \left[ x \right] = t = \left\\{ X \right\\} (Fractional part)
$\therefore$ $0 \leqslant t \leqslant 1$
Hence, $0 \leqslant \dfrac{{1 \pm \sqrt {1 + 3{a^2}} }}{3} \leqslant 1$ and $0 \leqslant \dfrac{{1 \mp \sqrt {1 + 3{a^2}} }}{3} \leqslant 1$
Taking only positive sign as x > 0 we get,
$0 \leqslant \dfrac{{1 + \sqrt {1 + 3{a^2}} }}{3} \leqslant 1$
Multiplying 3 with each we get,
$0 \times 3 \leqslant \dfrac{{1 + \sqrt {1 + 3{a^2}} }}{3} \times 3 < 1 \times 3$
I.e. $0 \leqslant 1 + \sqrt {1 + 3{a^2}} \leqslant 3$
Taking right side part only,
$1 + \sqrt {1 + 3{a^2}} \leqslant 3$
Subtracting 1 from each we get,
$\sqrt {1 + 3{a^2}} < 2$
Taking square on both side we get,
$1 + 3{a^2} < 4$
Subtracting 1 from both of the side we get,
$3{a^2} < 3$
Cancelling 3 from both of the side,
${a^2} < 1$
Taking 1 to the left hand side,
${a^2} - 1 < 0$
Expanding the above equation we get,
$(a - 1)(a + 1) < 0$
$\therefore$ $a \in ( - 1,1)$ , for no integer solution of a, we consider the interval $( - 1,0) \cup (0,1)$ .
Hence all possible values of a lie in the interval $( - 1,0) \cup (0,1)$ .

So, the correct answer is “Option A”.

Note: A quadratic equation is an equation of the second degree i.e. it contains at least one term which is square.The quadratic formula is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.In above question we took only positive sign because x is greater than 0.