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Question Number 114485 by mohammad17 last updated on 19/Sep/20

Answered by Dwaipayan Shikari last updated on 19/Sep/20

$$f\left(x\right)=\frac{1}{x^3-3x^2+2x}=\frac{1}{x(x-2)(x-1)}$$$$x\in R-\{0,1,2\}\left(Domain\right)$$

Commented by mohammad17 last updated on 19/Sep/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 19/Sep/20

$$f\left(x\right)=\frac{ax+b}{bx+a}$$$$f\left(x\right).f\left(\frac{1}{x}\right)=\frac{ax+b}{bx+a}.\frac{a+bx}{ax+b}=1$$$$$$$$3)f\left(x\right)=a^x$$$$Range=(0,+\mathrm{\infty })$$

Commented by mohammad17 last updated on 19/Sep/20

$$sir\left(3\right)trueorfalse?$$

Answered by 1549442205PVT last updated on 19/Sep/20

$$1)\mathrm{False}\mathrm{since}{\mathrm{x}}^3-{3\mathrm{x}}^2+2\mathrm{x}=0$$$$\iff \mathrm{x}(\mathrm{x}-1)(\mathrm{x}-2)=0\iff \mathrm{x}\in \{0,1,2\}$$$$\Rightarrow {\mathrm{D}}_{\mathrm{f}}=\mathrm{R}\mathrm{\setminus }\{0,1,2\}$$$$2)\mathrm{True}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{bx}+\mathrm{a}}\Rightarrow \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{a}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{b}}{\mathrm{b}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{a}}$$$$=\frac{\mathrm{a}+\mathrm{bx}}{\mathrm{b}+\mathrm{ax}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{bx}+\mathrm{a}}\times \frac{\mathrm{a}+\mathrm{bx}}{\mathrm{b}+\mathrm{ax}}=1$$$$3)\mathrm{True}:\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}^{\mathrm{x}}\Rightarrow \mathrm{D}:\mathrm{R};\mathrm{Range}:(0;+\mathrm{\infty })$$$$4)\mathrm{True}:\mathrm{f}\left(\mathrm{x}\right)=-4\mathrm{x}-7\Rightarrow \mathrm{f}(-3)=-4(-3)-7=12-7=5$$$$5)\mathrm{True}:\mathrm{f}\left(\mathrm{x}\right)={\log }_{\mathrm{a}}\mathrm{x}\Rightarrow \mathrm{D}:(0;+\mathrm{\infty });$$$$\mathrm{Range}:(-\mathrm{\infty };+\mathrm{\infty })\mathrm{or}\mathrm{R}$$

Commented by mohammad17 last updated on 19/Sep/20

$$butsir\left(3\right)and\left(5\right)howistrue$$

Commented by mohammad17 last updated on 19/Sep/20

$$and6)trueorfalse$$

Commented by 1549442205PVT last updated on 19/Sep/20

$$\mathrm{available}\mathrm{answer}\mathrm{is}\mathrm{true}$$

Commented by 1549442205PVT last updated on 19/Sep/20

$$\mathrm{available}\mathrm{assertion}\mathrm{is}\mathrm{true}.\mathrm{It}\mathrm{is}\mathrm{a}$$$$\mathrm{rational}\mathrm{function}$$

Commented by PRITHWISH SEN 2 last updated on 19/Sep/20

$$\mathrm{I}\mathrm{think}\mathrm{there}\mathrm{should}\mathrm{be}\mathrm{a}\mathrm{restriction}\mathrm{on}6\mathrm{and}\mathrm{it}\mathrm{is}$$$$\mid \mathbf{x}\mid \ne 2$$

Commented by Aziztisffola last updated on 20/Sep/20

$$\left(3\right)\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{xlna}}\Rightarrow \mathrm{D}=\mathbb{R}$$$$\underset{x\to -\mathrm{\infty }}{\lim f}\left(x\right)=0\mathrm{\&}\underset{x\to \mathrm{\infty }}{\lim f}\left(\mathrm{x}\right)=+\mathrm{\infty }$$$$\Rightarrow \mathrm{R}=(0;\mathrm{\infty })\mathrm{then}\mathrm{True}.$$

Commented by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{Question}6{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{require}\mathrm{to}\mathrm{find}\mathrm{domain}$$$$\mathrm{but}\mathrm{to}\mathrm{assert}\mathrm{the}\mathrm{true}\mathrm{or}\mathrm{the}\mathrm{false}$$

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