JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 18)
Let [t] denote the greatest integer $$ \le $$ t. Then the equation in x,
[x]2 + 2[x+2] - 7 = 0 has :
[x]2 + 2[x+2] - 7 = 0 has :
no integral solution.
exactly two solutions.
exactly four integral solutions.
infinitely many solutions.
Explanation
$${[x]^2} + 2[x + 2] - 7 = 0$$
$$ \Rightarrow $$ $${[x]^2} + 2[x] + 4 - 7 = 0$$
Using the property [x + n] = [x] + n ; n $$ \in $$ I
$$ \Rightarrow $$ $${[x]^2} + 2[x] - 3 = 0$$
let [x] = y
$${y^2} + 3y - y - 3 = 0$$
$$ \Rightarrow $$ $$(y - 1)(y + 3) = 0$$
$$[x] = 1\,or\,[x] = - 3$$
$$ \therefore $$ $$x \in \left[ {1,2} \right)\,\& \, \in \left[ { - 3, - 2} \right)$$
$$ \Rightarrow $$ $${[x]^2} + 2[x] + 4 - 7 = 0$$
Using the property [x + n] = [x] + n ; n $$ \in $$ I
$$ \Rightarrow $$ $${[x]^2} + 2[x] - 3 = 0$$
let [x] = y
$${y^2} + 3y - y - 3 = 0$$
$$ \Rightarrow $$ $$(y - 1)(y + 3) = 0$$
$$[x] = 1\,or\,[x] = - 3$$
$$ \therefore $$ $$x \in \left[ {1,2} \right)\,\& \, \in \left[ { - 3, - 2} \right)$$
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