If the domain of the function f(x) = log_{(10 | vedclass

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(D) For the function $f(x) = \log_{g(x)}h(x)$ to be defined, we need $h(x) > 0$, $g(x) > 0$, and $g(x) 
eq 1$. Step $1$: $h(x) = 18x^2 - 11x + 1 > 0

Vedclass · Accepted Answer

$316$

Vedclass · Answer

(D) For the function $f(x) = \log_{g(x)}h(x)$ to be defined, we need $h(x) > 0$, $g(x) > 0$, and $g(x) 
eq 1$. Step $1$: $h(x) = 18x^2 - 11x + 1 > 0 \implies (2x-1)(9x-1) > 0 \implies x  \frac{1}{2}$. Step $2$: $g(x) = 10x^2 - 17x + 7 > 0 \implies (x-1)(10x-7) > 0 \implies x  1$. Step $3$: $g(x) 
eq 1 \implies 10x^2 - 17x + 6 
eq 0 \implies (2x-1)(5x-6) 
eq 0 \implies x 
eq \frac{1}{2}, x 
eq \frac{6}{5}$. Combining these conditions: $x \in (-\infty, \frac{1}{9}) \cup (\frac{1}{2}, \frac{7}{10}) \cup (1, \infty) - \{\frac{6}{5}\}$. Comparing with $(-\infty, a) \cup (b, c) \cup (d, \infty) - \{e\}$, we get $a = \frac{1}{9}, b = \frac{1}{2}, c = \frac{7}{10}, d = 1, e = \frac{6}{5}$. Then $90(a+b+c+d+e) = 90(\frac{1}{9} + \frac{1}{2} + \frac{7}{10} + 1 + \frac{6}{5}) = 10 + 45 + 63 + 90 + 108 = 316$.

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:

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