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Question Number 208453 by hardmath last updated on 16/Jun/24

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)=\frac{(2\mathrm{a}+1)\cdot \mathrm{x}+1}{\mathrm{x}-\mathrm{a}}\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$\mathrm{Find}:{\mathrm{a}}^2+3=?$$

Answered by efronzo1 last updated on 16/Jun/24

$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{ax}+1}{\mathrm{x}-(2\mathrm{a}+1)}=\frac{(2\mathrm{a}+1)\mathrm{x}+1}{\mathrm{x}-\mathrm{a}}$$$$(\mathrm{ax}+1)(\mathrm{x}-\mathrm{a})=(\mathrm{x}-(2\mathrm{a}+1))((2\mathrm{a}+1)\mathrm{x}+1)$$$${\mathrm{ax}}^2-({\mathrm{a}}^2+1)\mathrm{x}-\mathrm{a}=(2\mathrm{a}+1){\mathrm{x}}^2+(1-{(2\mathrm{a}+1)}^2)\mathrm{x}-(2\mathrm{a}+1)$$$$\Rightarrow \mathrm{a}=2\mathrm{a}+1;\mathrm{a}=-1$$$$\Rightarrow {\mathrm{a}}^2+3=4$$

Answered by MM42 last updated on 16/Jun/24

$$f\left(f\left(0\right)\right)=0\Rightarrow f(-\frac{1}{a})=0$$$$\Rightarrow \frac{(2a+1)(-\frac{1}{a})+1}{-\frac{1}{a}-a}=0$$$$\Rightarrow \frac{2a+1}{a}=1\Rightarrow a=-1\Rightarrow a^2+3=4✓$$

Commented by hardmath last updated on 16/Jun/24

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professor}$$

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