If domain of the function f(x) 1[x]2−[x]−2 is (Here [x] denotes the greatest integer notexceeding the value of x)

f(x)=1[x]2−[x]−2 is defined iff[x]2−[x]−2> 0⇒ [x]2−2[x]+[x]−2> 0⇒([x]−2)([x]+1)> 0∴[x]+1 0⇒x 3∴x∈(−∞, −1) ∪ (3, ∞)

If domain of the function f(x) 1[x]2−[x]−2 is (Here [x] denotes the greatest integer notexceeding the value of x)

f(x)=1[x]2−[x]−2 is defined iff[x]2−[x]−2> 0⇒ [x]2−2[x]+[x]−2> 0⇒([x]−2)([x]+1)> 0∴[x]+1 0⇒x 3∴x∈(−∞, −1) ∪ (3, ∞)

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$I f d o m a i n o f t h e f u n c t i o n f (x) \frac{1}{\sqrt{{[x]}^{2} - [x] - 2}} i s (H e r e [x] d e n o t e s t h e g r e a t e s t int e g e r n o t e x c e e d i n g t h e v a l u e o f x)$

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

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

$(- \infty, - 2) \cup (0, \infty)$

$(- \infty, - 1) \cup [3, \infty)$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{\sqrt{{[x]}^{2} - [x] - 2}} i s d e f i n e d i f f \\ {[x]}^{2} - [x] - 2 > 0 \\ \Rightarrow {[x]}^{2} - 2 [x] + [x] - 2 > 0 \\ \Rightarrow ([x] - 2) ([x] + 1) > 0 \\ ∴ [x] + 1 < 0 o r [x] - 2 > 0 \\ \Rightarrow x < - 1 o r x > 3 \\ ∴ x \in (- \infty, - 1) \cup (3, \infty) \end{array}$