Question

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

• A. 316
• B. 320
• C. 163
• D. 631

Hint:

For logarithmic domains, always check three conditions carefully: base positive, base not equal to 1, and argument positive.

Answer 1

The Correct Option is A

Step 1: Write the conditions for logarithmic function.
For $f(x)=\log_{g(x)}(h(x))$ to be defined: \[ g(x)>0,\quad g(x)\neq 1,\quad h(x)>0 \] Here, \[ g(x)=10x^2-17x+7,\quad h(x)=18x^2-11x+1 \] Step 2: Solve $h(x)>0$. 
\[ 18x^2-11x+1>0 \] Roots: \[ x=\frac{1}{9},\; \frac{1}{2} \] Since the coefficient of $x^2$ is positive: \[ x<\frac{1}{9}\quad \text{or}\quad x>\frac{1}{2} \] Step 3: Solve $g(x)>0$. 
\[ 10x^2-17x+7>0 \] Roots: \[ x=\frac{7}{10},\;1 \] Hence, \[ x<\frac{7}{10}\quad \text{or}\quad x>1 \] Step 4: Exclude $g(x)=1$. 
\[ 10x^2-17x+7=1 \Rightarrow 10x^2-17x+6=0 \] Roots: \[ x=\frac{3}{5},\;1 \] These values must be excluded from the domain. Step 5: Combine all conditions. 
Intersecting all valid intervals: \[ (-\infty,\tfrac{1}{9})\cup(\tfrac{1}{2},\tfrac{3}{5})\cup(\tfrac{7}{10},\infty)-\{1\} \] Thus, \[ a=\frac{1}{9},\quad b=\frac{1}{2},\quad c=\frac{3}{5},\quad d=\frac{7}{10},\quad e=1 \] Step 6: Compute the required value. 
\[ 90(a+b+c+d+e) =90\left(\frac{1}{9}+\frac{1}{2}+\frac{3}{5}+\frac{7}{10}+1\right) =316 \]