| (i) \[f\left( x \right)\] =sgn\[({{x}^{3}}-x)\] |
| (ii) \[f\left( x \right)\] =sgn \[(2\cos x-1)\] |
| (iii) \[f\left( x \right)\] =sgn \[({{x}^{2}}-2x+3)\] |
A) Only (i)
B) Only (iii)
C) Both (ii) and (iii)
D) None of these
Correct Answer: B
Solution :
[b] (i) \[f(x)=\sgn ({{x}^{3}}-x)\] Here \[{{x}^{3}}-x=0\Rightarrow x=0,-1,1\] Hence, \[f(x)\] is discontinuous at \[x=0,-1,1\]. (ii) If \[(x)=\sgn (2\,\,\cos \,x-1)\] Here, \[2\cos \,\,x-1=0\] \[\Rightarrow \cos \,\,x=1/2\Rightarrow x=2n\pi +(\pi /3)\] \[n\in Z,\] where \[f(x)\] is discontinuous. (iii) \[f(x)=\sgn ({{x}^{2}}-2x+3)\] Here, \[{{x}^{2}}-2x+3>0\] for all x. Thus, \[f(x)=1\] for all x Hence continuous for all x.
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