Question Number 33818 by rahul 19 last updated on 25/Apr/18
$$\boldsymbol{{L}}{et}\:{f}:\boldsymbol{{R}}\:\rightarrow\:\left[\:\mathrm{1},\:\infty\right)\:{be}\:{defined}\:{as}\: \\ $$$${f}\left({x}\right)\:=\:\mathrm{log}_{\mathrm{10}} \:\left(\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\boldsymbol{{k}}+\mathrm{1}}\:+\mathrm{10}\:\right). \\ $$$$\boldsymbol{{I}}{f}\:{f}\left({x}\right)\:{is}\:\boldsymbol{{surjective}}\:,\:{then}\:{find} \\ $$$${the}\:{value}\:{of}\:\boldsymbol{{k}}\:? \\ $$
Answered by MJS last updated on 26/Apr/18
$$\mathrm{this}\:\mathrm{one}'\mathrm{s}\:\mathrm{easier} \\ $$$$\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+\boldsymbol{{k}}+\mathrm{1}}\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{must}\:\mathrm{have}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{zero} \\ $$$$\left(\mathrm{2}\:\mathrm{distinct}\:\mathrm{zeros}\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\right. \\ $$$$\left.\mathrm{some}\:{x}\in\mathbb{R};\:\mathrm{no}\:\mathrm{zeros}\:\Rightarrow\:{min}\left({f}\left({x}\right)\right)>\mathrm{1}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}+{k}+\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}{x}+\frac{{k}+\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}\pm\frac{\sqrt{\mathrm{1}−\mathrm{3}{k}}}{\mathrm{3}} \\ $$$$\mathrm{1}−\mathrm{3}{k}=\mathrm{0}\:\Rightarrow\:{k}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by rahul 19 last updated on 28/Apr/18
$${pls}\:\:{explain}\:{one}\:{more}\:{time}. \\ $$$${If}\:{f}\left({x}\right)\:{has}\:{exactly}\:{one}\:{zero}\:{then} \\ $$$${it}\:{will}\:{not}\:{be}\:{defined}\:{for}\:{that}\:{x}\:{i}.{e} \\ $$$${will}\:{not}\:{be}\:{sujective}\:…..\:?? \\ $$$${for}\:{eg}:\:{if}\:{it}\:{becomes}\:\left({x}−\mathrm{2}\right)\: \\ $$$${then}\:{f}\left({x}\right)\:{is}\:{not}\:{defined}\:{at}\:{x}=\mathrm{2}\:? \\ $$
Commented by MJS last updated on 28/Apr/18
![not f(x), I only talk about the root f(x)=log_(10) ((√(g(x)))+10) g(x) has { ((no zero ⇒ (√(g(x)))>0 ∧ f(x)>1 ∀x∈R ⇒)),((⇒ ∃y∈[1; ∞[: y≠f(x)∀x∈R)),((1 zero p ⇒ (√(g(p)))=0 ⇒ f(p)=1 ⇒)),((⇒ ∀y∈[1; ∞[: ∃x∈R: y=f(x) ⇒ f(x) is surjective)),((2 zeros p, q ⇒ (√(g(x)))∉R ∀x∈]p; q[ ⇒)),((⇒ D=R\x≠R ∀x∈]p; q[)) :}](https://www.tinkutara.com/question/Q33959.png)
$$\mathrm{not}\:{f}\left({x}\right),\:\mathrm{I}\:\mathrm{only}\:\mathrm{talk}\:\mathrm{about}\:\mathrm{the}\:\mathrm{root} \\ $$$${f}\left({x}\right)=\mathrm{log}_{\mathrm{10}} \:\left(\sqrt{{g}\left({x}\right)}+\mathrm{10}\right) \\ $$$${g}\left({x}\right)\:\mathrm{has}\:\begin{cases}{\mathrm{no}\:\mathrm{zero}\:\Rightarrow\:\sqrt{{g}\left({x}\right)}>\mathrm{0}\:\wedge\:{f}\left({x}\right)>\mathrm{1}\:\forall{x}\in\mathbb{R}\:\Rightarrow}\\{\Rightarrow\:\exists{y}\in\left[\mathrm{1};\:\infty\left[:\:{y}\neq{f}\left({x}\right)\forall{x}\in\mathbb{R}\right.\right.}\\{\mathrm{1}\:\mathrm{zero}\:{p}\:\Rightarrow\:\sqrt{{g}\left({p}\right)}=\mathrm{0}\:\Rightarrow\:{f}\left({p}\right)=\mathrm{1}\:\Rightarrow}\\{\Rightarrow\:\forall{y}\in\left[\mathrm{1};\:\infty\left[:\:\exists{x}\in\mathbb{R}:\:{y}={f}\left({x}\right)\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{surjective}\right.\right.}\\{\left.\mathrm{2}\:\mathrm{zeros}\:{p},\:{q}\:\Rightarrow\:\sqrt{{g}\left({x}\right)}\notin\mathbb{R}\:\forall{x}\in\right]{p};\:{q}\left[\:\Rightarrow\right.}\\{\left.\Rightarrow\:\mathbb{D}=\mathbb{R}\backslash{x}\neq\mathbb{R}\:\forall{x}\in\right]{p};\:{q}\left[\right.}\end{cases} \\ $$
Commented by rahul 19 last updated on 29/Apr/18
$$\mathscr{T}{hank}\:{you}\:{sir}. \\ $$