The domain of the real-valued function f(x) = | vedclass

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

Question

(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:$1$. The expression inside the square root in the numerator must

Vedclass · Accepted Answer

$[4, \infty)$

Vedclass · Answer

(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:$1$. The expression inside the square root in the numerator must be non-negative: $\log_{10}\left(\frac{x}{x-2}\right) \geq 0$.This implies $\frac{x}{x-2} \geq 10^0$,so $\frac{x}{x-2} \geq 1$.$\frac{x}{x-2} - 1 \geq 0$ $\Rightarrow \frac{x - (x-2)}{x-2} \geq 0$ $\Rightarrow \frac{2}{x-2} > 0$.This holds when $x-2 > 0$,i.e.,$x > 2$.$2$. The expression inside the square root in the denominator must be strictly positive: $[x]^2 - 5[x] + 6 > 0$.$([x]-2)([x]-3) > 0$.This implies $[x]  3$.If $[x]  3$,then $x \geq 4$.$3$. Combining the conditions $x > 2$ and ($x < 2$ or $x \geq 4$),we get $x \in [4, \infty)$.

The domain of the real-valued function $f(x) = \frac{\sqrt{\log_{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2-5[x]+6}}$ is (where $[x]$ denotes the greatest integer function):

Similar Questions

Range of the function $f(x) = 3 + 2^x + 4^x$ is

The domain and range of $f(x) = \frac{1}{\sqrt{|x| - x^2}}$ are $A$ and $B$ respectively. Then $A \cup B =$

The range of the real valued function $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$ is

The range of the real valued function $f(x) = \frac{15}{3 \sin x + 4 \cos x + 10}$ is

Given that $a, b$ and $c$ are real numbers such that $b^2 = 4ac$ and $a > 0$. The maximal possible set $D \subseteq R$ on which the function $f: D \rightarrow R$ given by $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$ is defined,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module