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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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$- 5 \leq x < - 3$

$1 \leq x < 4$

$3 \leq x < 5$

$2 \leq x < 4$

answer is B.

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Detailed Solution

$\begin{array}{l} {[x]}^{2} - 7 [x] + 12 \leq 0 \\ \Rightarrow {[x]}^{2} - 4 [x] - 3 [x] + 12 \leq 0 \\ \Rightarrow ([x] - 4) ([x] - 3) \leq 0 \\ \Rightarrow 3 \leq [x] \leq 4 \\ \Rightarrow x \in [3, 5) \end{array}$

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