Let f(x)=(x+1)2−1,(x≥−1). Then the set S={x:f(x)=f−1(x)}. is
A
Empty
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B
{0,−1}
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C
{0,1,−1}
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D
{0,−1,−3+i√32,−3−i√32}
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Solution
The correct option is D{0,−1,−3+i√32,−3−i√32} Letf(x)=(x+1)2−1,x≥−1.Sincef(x)=f−1(x)∴(x+1)2−1=√1+x−1(∵f−1(x)=√1+x−1)⇒(x+1)4=1+x⇒(x+1)[(x+1)3−1]=0⇒x=−1or(x+1)3=1⇒x+1=1,ω,ω2⇒x=0,−1,−3+i√32,−3−i√32