The domain of f(x) = [sin x] cos left( frac{p | vedclass

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

Question

(D) For the function $f(x) = [\sin x] \cos \left( \frac{\pi}{[x - 1]} \right)$ to be defined,the denominator inside the cosine function must not be ze

Vedclass · Accepted Answer

$R - [1, 2)$

Vedclass · Answer

(D) For the function $f(x) = [\sin x] \cos \left( \frac{\pi}{[x - 1]} ight)$ to be defined,the denominator inside the cosine function must not be zero.Thus,$[x - 1] 
eq 0$.By the definition of the Greatest Integer Function,$[x - 1] = 0$ when $0 \leq x - 1 Adding $1$ to all parts of the inequality,we get $1 \leq x Therefore,the function is undefined for $x \in [1, 2)$.The domain of the function is all real numbers except the interval $[1, 2)$,which is written as $R - [1, 2)$.

The domain of $f(x) = [\sin x] \cos \left( \frac{\pi}{[x - 1]} \right)$ is (where $[.]$ denotes the Greatest Integer Function $G.I.F.$).

Similar Questions

If $f: R \rightarrow R$ is defined by $f(x) = \frac{1}{2 - \cos 3x}$ for each $x \in R$,then the range of $f$ is

The domain of the function $f(x) = \sqrt{\log_{10}\left(\frac{5x - x^2}{4}\right)}$ is:

If $f:R \to S$ defined by $f(x) = \sin x - \sqrt{3} \cos x + 1$ is onto,then the interval of $S$ is

Find the domain of the function $f(x) = \frac{1}{\sqrt{(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6)}}$.

The domain of the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module