(A) 170

(B) 316

(C) 177

(D) 307

Correct Answer: 316

Explanation

For the logarithmic function to be defined:

Base must satisfy:

$$ 10x^2-17x+7>0 \quad \text{and} \quad 10x^2-17x+7\ne1 $$

Argument must satisfy:

$$ 18x^2-11x+1>0 $$

Solving base inequality:

$$ 10x^2-17x+7=0 $$

$$ (5x-7)(2x-1)=0 $$

$$ x=\frac{1}{2},\;\frac{7}{5} $$

Since coefficient of $x^2$ is positive,

$$ 10x^2-17x+7>0 \Rightarrow x<\frac12 \;\text{or}\; x>\frac75 $$

Solving argument inequality:

$$ 18x^2-11x+1=0 $$

$$ (9x-1)(2x-1)=0 $$

$$ x=\frac19,\;\frac12 $$

Thus,

$$ 18x^2-11x+1>0 \Rightarrow x<\frac19 \;\text{or}\; x>\frac12 $$

Base not equal to 1:

$$ 10x^2-17x+7=1 $$

$$ 10x^2-17x+6=0 $$

$$ (5x-6)(2x-1)=0 $$

$$ x=\frac12,\;\frac65 $$

$x=\frac12$ is already excluded; additional exclusion is

$$ x=\frac65 $$

Combining all conditions, domain becomes:

$$ (-\infty,\tfrac19)\cup(\tfrac12,\tfrac65)\cup(\tfrac75,\infty) $$

Thus,

$$ a=\frac19,\; b=\frac12,\; c=\frac65,\; d=\frac75,\; e=\frac65 $$

$$ a+b+c+d+e=\frac19+\frac12+\frac65+\frac75+\frac65=\frac{158}{45} $$

$$ 90(a+b+c+d+e)=90\times\frac{158}{45}=316 $$

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