f:R→R is defined by f(x)=[10−|x|] where [.] is the G.I.F then {x∈R/ f(x)=7} is

f(x)=[10− |x|]Given f(x)=7⇒ [10− |x|] =7⇒ 7≤10−|x| 2Let |x| ≤3, and |x| >2⇒ −3≤x≤3 and x 2x∈ [−3, −2)(2, 3]

f:R→R is defined by f(x)=[10−|x|] where [.] is the G.I.F then {x∈R/ f(x)=7} is

f(x)=[10− |x|]Given f(x)=7⇒ [10− |x|] =7⇒ 7≤10−|x| 2Let |x| ≤3, and |x| >2⇒ −3≤x≤3 and x 2x∈ [−3, −2)(2, 3]

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$f : R \to R i s d e f i n e d b y f (x) = [10 - | x |] w h e r e [.] i s t h e G . I . F t h e n {x \in R / f (x) = 7} i s$

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$[- 3, - 2]$

$[- 3, - 2) U (2, 3]$

$[- 3, 3]$

$[3, 7]$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = [10 - | x |] \\ G i v e n f (x) = 7 \\ \Rightarrow [10 - | x |] = 7 \\ \Rightarrow 7 \leq 10 - | x | < 8 \\ \Rightarrow - 10 + 7 \leq - | x | < 8 \\ \Rightarrow - 3 \leq - | x | < - 2 \\ \Rightarrow 3 \geq | x | > 2 \\ L e t | x | \leq 3, a n d | x | > 2 \\ \Rightarrow - 3 \leq x \leq 3 a n d x < - 2 a n d x > 2 \\ x \in [- 3, - 2) (2, 3] \end{array}$