ISI2015-DCG-49

Question

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

• A. set of positive real numbers
• B. set of real numbers
• C. set of negative real numbers
• D. set of real numbers larger than $5$

Answer 1

Answer $(A)$ The domain of $3x^2-4x+5$ is the ser to all Real numbers, $R$

But we know that the value of $\bf{\log}$ should be positive.

So the Domain finally becomes the set of all positive real numbers, i.e. $R^+$

 

If someone is more interested in finding the range also.

Then it comes out to be $\Big[\ln\frac{11}{3}, \infty\Big]$

It can be calculated this way.

The quadratic equation given forms the upward open parabola.

So, we just need to calculate its minimum value and the maximum value it can take is $\infty$.

So, the minimum value is calculated this way

Discriminant of the above equation is $D = b^2-4ac = \frac{16-4(3)(5)}{4(3)} = \frac{-44}{12} = \frac{11}{3}$

So $\bf{RANGE} = \Big[\ln\frac{11}{3}, \infty\Big]$

Comments

@`JEET

Why negative real numbers are not included in domain??

Because on substitution of a negative number, every term of $3x^{2}-4x+5$ becomes positive

Also there is no possibility of $3x^{2}-4x+5$ becoming 0 because it's roots are imaginary.

So i think domain must be set of all real numbers..

Correct me if iam wrong...

---

@chirudeepnamini

I think You are right.

The domain of $:\ln(3x^{2}−4x+5)$

Definition of logarithm$:3x^{2}−4x+5>0\:\: ,x \in \mathbf{R}$

Answer 2

All real numbers, even 0

Because the quadratic function inside the ln is a parabola upwards with a > 0 and D < 0, so no real roots, i.e ln(0) will never comes.

Also, the parabola always lies above x- axis, so, for all x, y will be positive. Hence range is R

Option B.