The domain of the function f(x)=loge sgn 9-x2+[x]3-4[x] where [.] = G.I.F

Given f(x)=loge sgn 9-x2+[x]3-4[x]=y1+y2(say) Now, y1 is defined if sgn9-x2>0But sgn x=1(i.e. > 0) if x>0∴sgn 9-x2>0⇒9-x2>0⇒x2-9<0⇒(x-3)(x+3)<0⇒-3<x<3Again y2 is defined if [x]3-4[x]≥0⇒[x][x]2-4≥0⇒[x]([x]-2)([x]+2)≥0Following the wavy curve method, we findThus [x]≥2 or [x] lies between – 2 and 0, i.e.[x] = -2, -1 or 0Now, [x]≥2⇒x≥2[x]=-2⇒-2≤x<1[x]=-1⇒-1≤x<0[x]=0⇒0≤x<1Hence [x]=-2,-1,0⇒-2≤x<1∴(B)∪(C)=(x≥2)...(C)Hence Df=(A)∪(C)=[-2,1)∪[2,3)

The domain of the function f(x)=loge sgn 9-x2+[x]3-4[x] where [.] = G.I.F

Given f(x)=loge sgn 9-x2+[x]3-4[x]=y1+y2(say) Now, y1 is defined if sgn9-x2>0But sgn x=1(i.e. > 0) if x>0∴sgn 9-x2>0⇒9-x2>0⇒x2-9<0⇒(x-3)(x+3)<0⇒-3<x<3Again y2 is defined if [x]3-4[x]≥0⇒[x][x]2-4≥0⇒[x]([x]-2)([x]+2)≥0Following the wavy curve method, we findThus [x]≥2 or [x] lies between – 2 and 0, i.e.[x] = -2, -1 or 0Now, [x]≥2⇒x≥2[x]=-2⇒-2≤x<1[x]=-1⇒-1≤x<0[x]=0⇒0≤x<1Hence [x]=-2,-1,0⇒-2≤x<1∴(B)∪(C)=(x≥2)...(C)Hence Df=(A)∪(C)=[-2,1)∪[2,3)

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The domain of the function $f (x) = \log_{e} \{sgn (9 - x^{2})\} + \sqrt{[x]^{3} - 4 [x]}$ where [.] = G.I.F

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$[- 2,1) \cup [2,3)$

$[- 4,1) \cup [2,3)$

$[4,1) \cup [2,3)$

$[2,1) \cup [2,3)$

answer is A.

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Detailed Solution

Given $f (x) = \log_{e} \{sgn (9 - x^{2})\} + \sqrt{[x]^{3} - 4 [x]} = y_{1} + y_{2}$ (say) Now, $y_{1}$ is defined if sgn $(9 - x^{2}) > 0$

But $sgn x = 1 (i . e . > 0) if x > 0$

$∴ sgn (9 - x^{2}) > 0 \Rightarrow 9 - x^{2} > 0 \Rightarrow x^{2} - 9 < 0 \Rightarrow (x - 3) (x + 3) < 0 \Rightarrow - 3 < x < 3$ Again $y_{2}$ is defined if $[x]^{3} - 4 [x] \geq 0 \Rightarrow [x] \{[x]^{2} - 4\} \geq 0 \Rightarrow [x] ([x] - 2) ([x] + 2) \geq 0$ Following the wavy curve method, we find

Thus $[x] \geq 2$ or [x] lies between – 2 and 0, i.e.[x] = -2, -1 or 0
Now, $[x] \geq 2 \Rightarrow x \geq 2$

$[x] = - 2 \Rightarrow - 2 \leq x < 1 [x] = - 1 \Rightarrow - 1 \leq x < 0 [x] = 0 \Rightarrow 0 \leq x < 1$ Hence $[x] = - 2, - 1,0 \Rightarrow - 2 \leq x < 1 ∴ (B) \cup (C) = (x \geq 2)$ ...(C)Hence $D_{f} = (A) \cup (C) = [- 2,1) \cup [2,3)$