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Question Number 21168 by Tinkutara last updated on 15/Sep/17

$$\mathrm{Let}f\left(x\right)=\mid x-1\mid +\mid x-2\mid +\mid x-3\mid ,$$$$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}k\mathrm{for}\mathrm{which}f\left(x\right)$$$$=k\mathrm{has}$$$$1.\mathrm{no}\mathrm{solution}$$$$2.\mathrm{only}\mathrm{one}\mathrm{solution}$$$$3.\mathrm{two}\mathrm{solutions}\mathrm{of}\mathrm{same}\mathrm{sign}$$$$4.\mathrm{two}\mathrm{solutions}\mathrm{of}\mathrm{opposite}\mathrm{sign}$$

Answered by alex041103 last updated on 15/Sep/17

$$f\left(x\right)=\{\begin{array}{l}3x-6,x\in (3,\mathrm{\infty }),f\left(x\right)\in (3,\mathrm{\infty })\\ 3,x=3\\ x,x\in (2,3),f\left(x\right)\in (2,3)\\ 2,x=2\\ 4-x,x\in (1,2),f\left(x\right)\in (2,3)\\ 3,x=1\\ 6-3x,x\in (-\mathrm{\infty },1),f\left(x\right)\in (3,\mathrm{\infty })\end{array}$$$$\Rightarrow f\left(x\right)\in [2,\mathrm{\infty })$$$$$$$$\Rightarrow 1.k\in (-\mathrm{\infty },2)$$$$\Rightarrow 2.k=2$$$$\Rightarrow 3.k=3$$$$\Rightarrow 4.k\in [6,\mathrm{\infty })$$

Commented by Tinkutara last updated on 15/Sep/17

$$\mathrm{Answer}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{part}\mathrm{is}\mathrm{wrong}.$$

Commented by alex041103 last updated on 15/Sep/17

$$f\left(x\right)=\{\begin{array}{l}3x-6,x\in (3,\mathrm{\infty }),f\left(x\right)\in (3,\mathrm{\infty })\\ 3,x=3\\ x,x\in (2,3),f\left(x\right)\in (2,3)\\ 2,x=2\\ 4-x,x\in (1,2),f\left(x\right)\in (2,3)\\ 3,x=1\\ 6-3x,x\in (-\mathrm{\infty },1),f\left(x\right)\in (3,\mathrm{\infty })\end{array}$$$$Youcanseethatforx=1,3f\left(x\right)=k=3$$$$sincesgn\left(1\right)=sgn\left(3\right)amdbecause$$$$theintervalsinblueareopenthere{aren}^{\prime }t$$$$anymorexforwhichf\left(x\right)=3$$$$AmIwrong?$$

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