If n is an integer,the domain of the function | vedclass

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

Question

(B) For the function $f(x) = \sqrt{\sin 2x}$ to be defined,the expression under the square root must be non-negative: $\sin 2x \geq 0$. The sine funct

Vedclass · Accepted Answer

$[n\pi, n\pi + \frac{\pi}{2}]$

Vedclass · Answer

(B) For the function $f(x) = \sqrt{\sin 2x}$ to be defined,the expression under the square root must be non-negative: $\sin 2x \geq 0$. The sine function is non-negative in the first and second quadrants,i.e.,in the intervals $[2n\pi, 2n\pi + \pi]$. Therefore,$2n\pi \leq 2x \leq 2n\pi + \pi$. Dividing by $2$,we get $n\pi \leq x \leq n\pi + \frac{\pi}{2}$. Thus,the domain is $[n\pi, n\pi + \frac{\pi}{2}]$.

If $n$ is an integer,the domain of the function $\sqrt{\sin 2x}$ is

Similar Questions

Let $A = \{x \in R, x \neq 0, -4 \leq x \leq 4\}$ and $f: A \rightarrow R$ be defined by $f(x) = \frac{|x|}{x}$ for $x \in A$. Then, the range of $f$ is

If the domain of the function $f(x) = x^2 - 6x + 7$ is $(-\infty, \infty)$,then the range of the function is

If $D \subseteq R$ and $f: D \rightarrow R$ defined by $f(x) = \frac{x^2+x+a}{x^2-x+a}$ is a surjection, then '$a$' lies in the interval

If $f(x) = [x]^{2} - 5[x] + 6 = 0$,where $[x]$ denotes the greatest integer function,then $x \in$

Range of the function $f(x) = \frac{x^2+x+2}{x^2+x+1}, x \in R$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module