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If-f-x-2a-1-x-1-x-a-and-f-x-f-1-x-Find-a-2-3-




Question Number 208453 by hardmath last updated on 16/Jun/24
If   f(x) = (((2a + 1)∙x + 1)/(x − a))   and   f(x) = f^(−1) (x)  Find:   a^2  + 3 = ?
$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\left(\mathrm{2a}\:+\:\mathrm{1}\right)\centerdot\mathrm{x}\:+\:\mathrm{1}}{\mathrm{x}\:−\:\mathrm{a}}\:\:\:\mathrm{and}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{3}\:=\:? \\ $$
Answered by efronzo1 last updated on 16/Jun/24
 f^(−1) (x)=((ax+1)/(x−(2a+1))) = (((2a+1)x+1)/(x−a))   (ax+1)(x−a)= (x−(2a+1))((2a+1)x+1)    ax^2 −(a^2 +1)x−a=(2a+1)x^2 +(1−(2a+1)^2 )x−(2a+1)    ⇒a=2a+1 ; a=−1    ⇒a^2 +3 = 4
$$\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{ax}+\mathrm{1}}{\mathrm{x}−\left(\mathrm{2a}+\mathrm{1}\right)}\:=\:\frac{\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{a}} \\ $$$$\:\left(\mathrm{ax}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{a}\right)=\:\left(\mathrm{x}−\left(\mathrm{2a}+\mathrm{1}\right)\right)\left(\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{x}+\mathrm{1}\right) \\ $$$$\:\:\mathrm{ax}^{\mathrm{2}} −\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{x}−\mathrm{a}=\left(\mathrm{2a}+\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{1}−\left(\mathrm{2a}+\mathrm{1}\right)^{\mathrm{2}} \right)\mathrm{x}−\left(\mathrm{2a}+\mathrm{1}\right) \\ $$$$\:\:\Rightarrow\mathrm{a}=\mathrm{2a}+\mathrm{1}\:;\:\mathrm{a}=−\mathrm{1} \\ $$$$\:\:\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{3}\:=\:\mathrm{4} \\ $$
Answered by MM42 last updated on 16/Jun/24
f(f(0))=0⇒f(−(1/a))=0  ⇒(((2a+1)(−(1/a))+1)/(−(1/a)−a))=0  ⇒((2a+1)/a)=1⇒a=−1⇒a^2 +3=4 ✓
$${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{0}\Rightarrow{f}\left(−\frac{\mathrm{1}}{{a}}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{\left(\mathrm{2}{a}+\mathrm{1}\right)\left(−\frac{\mathrm{1}}{{a}}\right)+\mathrm{1}}{−\frac{\mathrm{1}}{{a}}−{a}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{2}{a}+\mathrm{1}}{{a}}=\mathrm{1}\Rightarrow{a}=−\mathrm{1}\Rightarrow{a}^{\mathrm{2}} +\mathrm{3}=\mathrm{4}\:\checkmark \\ $$
Commented by hardmath last updated on 16/Jun/24
thankyou dear professor
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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