Question

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

• A. 177
• B. 170
• C. 307
• D. 316

Hint:

For logarithmic domains, always solve inequality by sign chart of the argument.

Answer 1

The Correct Option is A

Step 1: Condition for logarithm.
For $f(x)$ to be defined, \[ (10x^2-17x+7)(18x^2-11x+1)>0 \] Step 2: Finding roots. 
\[ 10x^2-17x+7=0 \Rightarrow x=1,\;\frac{7}{10} \] \[ 18x^2-11x+1=0 \Rightarrow x=\frac{1}{9},\;\frac{1}{2} \] Step 3: Sign analysis. 
Arranging roots: \[ \frac{1}{9}<\frac{1}{2}<\frac{7}{10}<1 \] The product is positive in intervals \[ (-\infty,\tfrac{1}{9})\cup(\tfrac{1}{2},\tfrac{7}{10})\cup(1,\infty) \] Thus, \[ a=\frac{1}{9},\quad b=\frac{1}{2},\quad c=\frac{7}{10},\quad d=1 \] The value excluded due to logarithm zero is \[ e=\frac{1}{2} \] Step 4: Final calculation. 
\[ a+b+c+d+e=\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{1}{2} =\frac{59}{30} \] \[ 90(a+b+c+d+e)=90\times\frac{59}{30}=177 \]