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(B) Given inequality is $[x]^2 - 7[x] + 12 \leq 0$. Let $y = [x]$. Then the inequality becomes $y^2 - 7y + 12 \leq 0$. Factoring the quadratic express

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$3 \leq x < 5$

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(B) Given inequality is $[x]^2 - 7[x] + 12 \leq 0$. Let $y = [x]$. Then the inequality becomes $y^2 - 7y + 12 \leq 0$. Factoring the quadratic expression,we get $(y - 4)(y - 3) \leq 0$. This implies that $3 \leq y \leq 4$. Since $y = [x]$,we have $3 \leq [x] \leq 4$. This means $[x]$ can be either $3$ or $4$. If $[x] = 3$,then $3 \leq x If $[x] = 4$,then $4 \leq x Combining these two intervals,we get $3 \leq x < 5$.

If $[x]$ denotes the greatest integer not exceeding $x$,then the values of $x$ satisfying $[x]^2-7[x]+12 \leq 0$ are

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