If the domain of the function f(x)= _(10x^2-17x+7)(18x^2-11x+1) is (-,a) (b,c) (d,)-e, the…

Question

If the domain of the function

$f(x)={\log }_{(10{x}^{2}-17x+7)}(18{x}^{2}-11x+1)$

is $(-\infty,a)\cup (b,c)\cup (d,\infty)-{e}$, then

$90(a+b+c+d+e)$ equals:

Accepted Answer

$\begin{gathered} f(x)={\log }_{(10{x}^{2}-17x+7)}(18{x}^{2}-11x+1) \ {\log }_{b}aa>0b>0b
e 1 \ 18{x}^{2}-11x+1>0 \ 18{x}^{2}-9x-2x+1>0 \ 9x(2x-1)-1(2x-1)>0 \ (2x-1)(9x-1)>0 \ {S}_{1}:x\in (-\infty,1/9)\cup (1/2,\infty) \ 10{x}^{2}-17x+7>0 \ 10{x}^{2}-10x-7x+7>0 \ 10x(x-1)-7(x-1)>0 \ (10x-7)(x-1)>0 \ {S}_{2}:x\in (-\infty,7/10)\cup (1,\infty) \ 10{x}^{2}-17x+7
e 1 \end{gathered}$

$\begin{gathered} 10{x}^{2}-17x+6
e 0 \ 10{x}^{2}-12x-5x+6
e 0 \ 2x(5x-6)-1(5x-6)
e 0 \ (2x-1)(5x-6)
e 0 \ {S}_{3}:x
e 1/2,6/5\Rightarrow x
e 0.5,1.2. \ x\in [{S}_{1}\cap {S}_{2}]-{0.5,1.2}. \ x\in [(-\infty,1/9)\cup (1/2,\infty)]\cap [(-\infty,7/10)\cup (1,\infty)]-{0.5,1.2}. \ x\in (-\infty,1/9)\cup (1/2,7/10)\cup (1,\infty)-{1.2}. \end{gathered}$

Since 0.5 is not included, it need not to be subtracted.

Compare final solution with $(-\infty,a)\cup (b,c)\cup (d,\infty)-{e}$.

$\begin{gathered} a=1/9,b=1/2,c=7/10,d=1e=1.2. \ 90(a+b+c+d+e)=90(1/9+1/2+7/10+1+1.2) \ =10+45+63+90+108=316. \end{gathered}$

Solutions · 1