Question:
Let \([t]\) denote the greatest integer \(\le t\). Then the equation in \(x\), \([x]^2+2[x+2]-7=0\) has :
Let \([t]\) denote the greatest integer \(\le t\). Then the equation in \(x\), \([x]^2+2[x+2]-7=0\) has :
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Use \([x+2]=[x]+2\) to get a quadratic in \([x]\); each integer value of \([x]\) corresponds to a whole interval of \(x\).
Updated On: Nov 5, 2026
- no integral solution.
- exactly two solutions.
- exactly four integral solutions.
- infinitely many solutions.
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The Correct Option is D
Solution and Explanation
\([x+2]=[x]+2\). Let \(n=[x]\): \(n^2+2(n+2)-7=0\Rightarrow n^2+2n-3=0\Rightarrow (n+3)(n-1)=0\).
So \([x]=1\) or \([x]=-3\), i.e. \(x\in[1,2)\) or \(x\in[-3,-2)\).
Both are intervals, so there are infinitely many solutions (the integral solutions are only \(x=1\) and \(x=-3\)).
Answer: (4) infinitely many solutions.
So \([x]=1\) or \([x]=-3\), i.e. \(x\in[1,2)\) or \(x\in[-3,-2)\).
Both are intervals, so there are infinitely many solutions (the integral solutions are only \(x=1\) and \(x=-3\)).
Answer: (4) infinitely many solutions.
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