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Question Number 208453 by hardmath last updated on 16/Jun/24
Iff(x)=(2a+1)⋅x+1x−aandf(x)=f−1(x)Find:a2+3=?
Answered by efronzo1 last updated on 16/Jun/24
f−1(x)=ax+1x−(2a+1)=(2a+1)x+1x−a(ax+1)(x−a)=(x−(2a+1))((2a+1)x+1)ax2−(a2+1)x−a=(2a+1)x2+(1−(2a+1)2)x−(2a+1)⇒a=2a+1;a=−1⇒a2+3=4
Answered by MM42 last updated on 16/Jun/24
f(f(0))=0⇒f(−1a)=0⇒(2a+1)(−1a)+1−1a−a=0⇒2a+1a=1⇒a=−1⇒a2+3=4✓
Commented by hardmath last updated on 16/Jun/24
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