Question Number 182657 by Strengthenchen last updated on 12/Dec/22
![as we can know for Q_1 , set the funvction in question as f(x) and make the first step as: a=1,f(x)=x^2 −7x+3ln x df(x)=2x−7+(3/x)=h(x) set h(x)=0⇒x=(1/2)&x=3 so,the monotonicity of f(x) is f(n)<f((1/2))_(∣n<(1/2)) ,f((1/2))>f(2),f(2)<f(p)_(∣p>2.) Q_1 has been proved finished Q_(2 ) ,set ax^2 −(a+6)x+3ln x>−6 when x∈[2,3e], make the f(x) dive two part as g(x)=ax^2 −(6+a)x&k(x)=−3(ln x+2), the middle value of g(x) is x=((a+6)/(2a)) when x=3e, k_(min) (3e)=−15, x=2, k_(max) (2)=−3(ln 2+2) g(2)=2a−12>k_(max) (x)⇒a>3−(3/2)ln 2 when x=(1/2)+(3/a), g(x)=−(((a+6)^2 )/(4a))>−3(ln (((a+6)/(2a)))+2)&a>0&2<(1/2)+(3/a)<3e](Q182657.png)
$${as}\:{we}\:{can}\:{know}\:{for}\:{Q}_{\mathrm{1}} ,\:{set}\:{the}\:{funvction}\:{in}\:{question}\:{as}\:{f}\left({x}\right) \\ $$ $${and}\:{make}\:{the}\:{first}\:{step}\:{as}: \\ $$ $${a}=\mathrm{1},{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{3ln}\:{x} \\ $$ $${df}\left({x}\right)=\mathrm{2}{x}−\mathrm{7}+\frac{\mathrm{3}}{{x}}={h}\left({x}\right) \\ $$ $${set}\:{h}\left({x}\right)=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\&{x}=\mathrm{3} \\ $$ $${so},{the}\:{monotonicity}\:{of}\:{f}\left({x}\right)\:{is}\:{f}\left({n}\right)<{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mid{n}<\frac{\mathrm{1}}{\mathrm{2}}} ,{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)>{f}\left(\mathrm{2}\right),{f}\left(\mathrm{2}\right)<{f}\left({p}\underset{\mid{p}>\mathrm{2}.} {\right)} \\ $$ $${Q}_{\mathrm{1}} \:{has}\:{been}\:{proved}\:{finished} \\ $$ $${Q}_{\mathrm{2}\:} \:,{set}\:{ax}^{\mathrm{2}} −\left({a}+\mathrm{6}\right){x}+\mathrm{3ln}\:{x}>−\mathrm{6}\:{when}\:{x}\in\left[\mathrm{2},\mathrm{3}{e}\right], \\ $$ $${make}\:{the}\:{f}\left({x}\right)\:{dive}\:{two}\:{part}\:{as} \\ $$ $$\:{g}\left({x}\right)={ax}^{\mathrm{2}} −\left(\mathrm{6}+{a}\right){x\&k}\left({x}\right)=−\mathrm{3}\left(\mathrm{ln}\:{x}+\mathrm{2}\right), \\ $$ $$\:{the}\:{middle}\:{value}\:{of}\:{g}\left({x}\right)\:{is}\:{x}=\frac{{a}+\mathrm{6}}{\mathrm{2}{a}} \\ $$ $${when}\:{x}=\mathrm{3}{e},\:{k}_{{min}} \left(\mathrm{3}{e}\right)=−\mathrm{15},\:\:{x}=\mathrm{2},\:{k}_{{max}} \left(\mathrm{2}\right)=−\mathrm{3}\left(\mathrm{ln}\:\mathrm{2}+\mathrm{2}\right) \\ $$ $${g}\left(\mathrm{2}\right)=\mathrm{2}{a}−\mathrm{12}>{k}_{{max}} \left({x}\right)\Rightarrow{a}>\mathrm{3}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$ $${when}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}},\:{g}\left({x}\right)=−\frac{\left({a}+\mathrm{6}\right)^{\mathrm{2}} }{\mathrm{4}{a}}>−\mathrm{3}\left(\mathrm{ln}\:\left(\frac{{a}+\mathrm{6}}{\mathrm{2}{a}}\right)+\mathrm{2}\right)\&{a}>\mathrm{0\&2}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}}<\mathrm{3}{e} \\ $$ $$ \\ $$