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The domain of the function $\ln(3x^{2}-4x+5)$ is

1. set of positive real numbers
2. set of real numbers
3. set of negative real numbers
4. set of real numbers larger than $5$

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Answer: $B$

Given: $$\ln \underbrace{(3x^2-4x+5)}_\text{f(x)}$$

Let $y =\ln \underbrace{(3x^2-4x+5)}_\text{f(x)}$

Let's suppose $f(x) = 3x^2 - 4x + 5$

Here, $a = 3, \;b = -4,\;and\;c = 5$

$Discriminant, D = b^2-4ac = -4^2-4.3.5 = 16-60 = -44 < 0$

Since, $D$ is negative, $\therefore$ $x$ can take any value in $\mathbb{R}$ which means it has no real roots $\implies$ For any value of $x$, $f(x)$ will always be $\bf{positive}$.

In $\ln f(x)$, $f(x)$ should always be "greater than 0" but as we have seen above $f(x)$ is always greater than 0 for any value of x.

$\therefore$ the Domain of $y$ is a set of all Real numbers, $\bf{\mathbb{R}}$

Hence, option $B$ is the correct answer.

by Boss (19.1k points)