If the domain of the function f(x) = _10x^2-17x+7(18x^2 - 11x + 1) is (-,a) (b,c) (d,) - e…

Question

If the domain of the function $f(x) = \log_{10x^2-17x+7}(18x^2 - 11x + 1)$ is $(-\infty,a) \cup (b,c) \cup (d,\infty) - \{e\}$, then $90(a+b+c+d+e)$ equals:

Accepted Answer

For $\log_b a$ to be defined we need $a>0$, $b>0$, $b\neq 1$.

Argument positive: $18x^2-11x+1>0 \Rightarrow (2x-1)(9x-1)>0 \Rightarrow x \in \left(-\infty,\tfrac19\right)\cup\left(\tfrac12,\infty\right)$. Call this $S_1$.

Base positive: $10x^2-17x+7>0 \Rightarrow (10x-7)(x-1)>0 \Rightarrow x \in \left(-\infty,\tfrac{7}{10}\right)\cup(1,\infty)$. Call this $S_2$.

Base $\ne 1$: $10x^2-17x+7 \ne 1 \Rightarrow 10x^2-17x+6 \ne 0 \Rightarrow (2x-1)(5x-6)\ne 0 \Rightarrow x \ne \tfrac12, \tfrac65$.

Combining $S_1 \cap S_2$ and removing $x=\tfrac65$ (since $x=\tfrac12$ is already excluded):

$$x \in \left(-\infty, \frac19\right) \cup \left(\frac12, \frac{7}{10}\right) \cup (1,\infty) - \left\{\frac65\right\}$$

So $a=\tfrac19,\ b=\tfrac12,\ c=\tfrac{7}{10},\ d=1,\ e=\tfrac65$.

$90(a+b+c+d+e) = 10+45+63+90+108 = 316$

Solutions · 1