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

Hint: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.