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if a\; \epsilon\; R  and the equation -3\left ( x-\left [ x \right ] \right )^{2}+2\left ( x-\left [ x \right ] \right )+a^{2}=0  (where \left [ x \right ]  denotes the greatest integer \leqslant x ) has no integral solution, then all possible values of a lie in the interval :

Option: 1

\left ( -2,-1 \right )\;


Option: 2

\left ( -\infty ,-2 \right )\cup \left ( 2,\infty \right )\;


Option: 3

\; \left ( -1,0 \right )\cup \left ( 0,1 \right )\; \;


Option: 4

\left ( 1,2 \right )


Answers (1)

best_answer

Here, we have,

-3\left ( x-\left [ x \right ] \right )^{2}+2\left ( x-\left [ x \right ] \right )+a^{2}=0

now, x - [x] = {x} = t (say) 

And, t \in [0, 1)

hence, we have 

-3t^2 + 2t +a^2 = 0 \\ \Rightarrow a^2 = 3t^2 - 2t = 3t(t-\frac{2}{3})

Now, Let us consider that this equation has some solution.

then, as LHS is always positive or zero.

And, t \in [0, 1)

Hence, RHS \in [0, 1]

therefore, for solution to happen LHS should also be same.

Hence, for no solution

a\in(-1,0)\cup(0,1)

Posted by

Ritika Kankaria

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