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Question Number 198367 by universe last updated on 18/Oct/23

$$\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\frac{x^2-(b+1)x+b}{x^2-(a+1)x+a}(a\ne b\mathrm{\&}a,b\in \mathbb{R}\sim \left\{1\right\})$$$$\mathrm{can}\mathrm{take}\mathrm{all}\mathrm{values}\mathrm{except}\mathrm{two}\mathrm{values}\alpha \mathrm{\&}\beta$$$$\mathrm{such}\mathrm{that}\alpha +\beta =0\mathrm{then}\mid \frac{{\mathrm{a}}^3+{\mathrm{b}}^3-8}{\mathrm{ab}}\mid =??$$

Commented by Frix last updated on 18/Oct/23

$$y=\frac{x^2-(b+1)x+b}{x^2-(a+1)x+a}=\frac{(x-1)(x-b)}{(x-1)(x-a)}=\frac{x-b}{x-a}$$$$x\ne 1\wedge x\ne a$$$$\iff$$$$x=\frac{ay-b}{y-1}\Rightarrow y\ne 1$$$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{only}1\mathrm{value}f\left(x\right)\mathrm{cannot}\mathrm{take}.$$$$\mathrm{If}\mathrm{you}\mathrm{mean}\mathrm{values}x\mathrm{cannot}\mathrm{take},\mathrm{these}$$$$\mathrm{are}1,a$$

Commented by universe last updated on 18/Oct/23

Commented by universe last updated on 18/Oct/23

$$answeris6$$

Commented by Frix last updated on 18/Oct/23

$$\mathrm{This}\mathrm{question}\mathrm{makes}\mathrm{no}\mathrm{sense}.$$

Commented by Tinku Tara last updated on 19/Oct/23

$$\mathrm{for}\mathrm{a}=\mathrm{b},y=1.\mathrm{So}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{case}\mathrm{when}\mathrm{y}=1$$

Commented by Frix last updated on 19/Oct/23

$$\mathrm{But}\mathrm{it}\mathrm{says}a\ne b$$

Commented by Tinku Tara last updated on 19/Oct/23

$$\mathrm{Correct},\mathrm{commented}\mathrm{without}$$$$\mathrm{readimg}\mathrm{question}$$

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