If [x] denotes the greatest integer ≤x, then the domain of the function f(x)= 4−x2[x]+2 is

f(x)=4−x2[x]+2 is defined iff4−x2[x]+2≥0,[x]+2≠0([x]+2)(4−x2)≥0,⇒[x]≠−2⇒([x]+2)(x2−4)≤ 0⇒([x]+2)(x+2)(x−2)≤0∴ x∈ (−∞, −2)∪[−1, 2]

If [x] denotes the greatest integer ≤x, then the domain of the function f(x)= 4−x2[x]+2 is

f(x)=4−x2[x]+2 is defined iff4−x2[x]+2≥0,[x]+2≠0([x]+2)(4−x2)≥0,⇒[x]≠−2⇒([x]+2)(x2−4)≤ 0⇒([x]+2)(x+2)(x−2)≤0∴ x∈ (−∞, −2)∪[−1, 2]

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$I f [x] d e n o t e s t h e g r e a t e s t int e g e r \leq x, t h e n t h e d o m a i n o f t h e f u n c t i o n f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}} i s$

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$(- \infty, - 2) (- \infty, - 2)$

$(- \infty, - 2) \cup [- 1, 2]$

$(- \infty, - 2) \cup (- 1, 2)$

$(- \infty, - 1] \cup [1, 2]$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\frac{4 - x^{2}}{[x] + 2}} i s d e f i n e d i f f \\ \frac{4 - x^{2}}{[x] + 2} \geq 0, [x] + 2 \neq 0 \\ ([x] + 2) (4 - x^{2}) \geq 0, \Rightarrow [x] \neq - 2 \\ \Rightarrow ([x] + 2) (x^{2} - 4) \leq 0 \\ \Rightarrow ([x] + 2) (x + 2) (x - 2) \leq 0 \\ ∴ x \in (- \infty, - 2) \cup [- 1, 2] \end{array}$