The domain of the function log _{10}(x^2-5x+6 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For the function $f(x) = \log _{10}(x^2-5x+6)$ to be defined,the argument of the logarithm must be strictly positive.$x^2-5x+6 > 0$Factorizing the

Vedclass · Accepted Answer

$(-\infty, 2) \cup (3, \infty)$

Vedclass · Answer

(B) For the function $f(x) = \log _{10}(x^2-5x+6)$ to be defined,the argument of the logarithm must be strictly positive.$x^2-5x+6 > 0$Factorizing the quadratic expression:$(x-2)(x-3) > 0$Using the wavy curve method or sign scheme,the expression is positive when $x$ lies outside the roots $2$ and $3$.Thus,$x  3$.Therefore,the domain is $x \in (-\infty, 2) \cup (3, \infty)$.

The domain of the function $\log _{10}(x^2-5x+6)$ is

Similar Questions

Domain of the function $f$,given by $f(x) = \frac{1}{\sqrt{(x - 2)(x - 5)}}$ is

The range of the function $f(x) = \sqrt{3-x} + \sqrt{2+x}$ is

The domain of the function $\sqrt{\log({x^2} - 6x + 6)}$ is

If function $f : R \to S, f(x) = (\sin x - \sqrt{3} \cos x + 1)$ is onto,then $S$ is equal to

The function $f: R \rightarrow R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$. Then $f(R)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module