If D is the domain and G is the range of the | vedclass

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

Question

(B) Given the function $f(x) = \sqrt{\frac{1-x^2}{1+x^2}}$.For the domain $D$, we require $\frac{1-x^2}{1+x^2} \geq 0$. Since $1+x^2 > 0$ for all real

Vedclass · Accepted Answer

$[0, 1]$

Vedclass · Answer

(B) Given the function $f(x) = \sqrt{\frac{1-x^2}{1+x^2}}$.For the domain $D$, we require $\frac{1-x^2}{1+x^2} \geq 0$. Since $1+x^2 > 0$ for all real $x$, we need $1-x^2 \geq 0$, which implies $x^2 \leq 1$, so $x \in [-1, 1]$. Thus, $D = [-1, 1]$.For the range $G$, let $y = \sqrt{\frac{1-x^2}{1+x^2}}$. Since $x \in [-1, 1]$, $1-x^2$ ranges from $0$ to $1$ and $1+x^2$ ranges from $1$ to $2$. Thus, $\frac{1-x^2}{1+x^2}$ ranges from $0$ to $1$. Taking the square root, $y \in [0, 1]$. Thus, $G = [0, 1]$.Finally, $D \cap G = [-1, 1] \cap [0, 1] = [0, 1]$.

If $D$ is the domain and $G$ is the range of the real-valued function $f(x)=\sqrt{\frac{1-x^2}{1+x^2}}$, then $D \cap G=$

Similar Questions

Find the domain and the range of the real function $f$ defined by $f(x) = |x - 1|$.

If the real valued function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is not defined for all $x \in(-\infty, a] \cup(b, \infty)$, then $3^a+b^2=$

Let the domain of the function $f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)$ be $[\alpha, \beta]$ and the domain of $g(x) = \log_2\left(2-6\log_{27}(2x+5)\right)$ be $(\gamma, \delta)$. Then $|7(\alpha+\beta)+4(\gamma+\delta)|$ is equal to . . . . . . .

Let $f:[2, \infty) \rightarrow R$ be the function defined by $f(x)=x^{2}-4x+5$. Then the range of $f$ is:

Is it true that $x = e^{\log x}$ for all real $x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module