If f: R rightarrow [-1, 1] and g: R rightarro | vedclass

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

Question

(A) Let $f(x) = y$. Since $f: R \rightarrow [-1, 1]$ is surjective,$y$ takes all values in $[-1, 1]$.Given $\sin \left(g(x) - \frac{\pi}{3}\right) = \

Vedclass · Accepted Answer

$\left[0, \frac{2 \pi}{3}\right]$

Vedclass · Answer

(A) Let $f(x) = y$. Since $f: R ightarrow [-1, 1]$ is surjective,$y$ takes all values in $[-1, 1]$.Given $\sin \left(g(x) - \frac{\pi}{3}ight) = \frac{y}{2} \sqrt{4 - y^2}$.Let $y = 2 \sin 	heta$. Since $y \in [-1, 1]$,$\sin 	heta \in \left[-\frac{1}{2}, \frac{1}{2}ight]$,so $	heta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}ight]$.Then $\frac{y}{2} \sqrt{4 - y^2} = \sin 	heta \sqrt{4 - 4 \sin^2 	heta} = \sin 	heta \sqrt{4 \cos^2 	heta} = 2 \sin 	heta \cos 	heta = \sin(2 	heta)$.Thus,$\sin \left(g(x) - \frac{\pi}{3}ight) = \sin(2 	heta)$.Since $	heta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}ight]$,$2 	heta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}ight]$.Therefore,$g(x) - \frac{\pi}{3} = 2 	heta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}ight]$.$g(x) \in \left[-\frac{\pi}{3} + \frac{\pi}{3}, \frac{\pi}{3} + \frac{\pi}{3}ight] = \left[0, \frac{2 \pi}{3}ight]$.Since $g$ is surjective,the codomain $A = \left[0, \frac{2 \pi}{3}ight]$.

If $f: R \rightarrow [-1, 1]$ and $g: R \rightarrow A$ are two surjective mappings and $\sin \left(g(x) - \frac{\pi}{3}\right) = \frac{f(x)}{2} \sqrt{4 - f^2(x)}$,then $A =$

Similar Questions

Let $[x]$ represent the greatest integer less than or equal to $x$,${x} = x - [x]$,$\sqrt{2} = 1.414$ and $\sqrt{3} = 1.732$. If $f(x) = \{x + [\frac{x}{1+x^2}]\}$ is a real-valued function,then $f(\sqrt{2}) + f(-\sqrt{3}) = $

The total number of functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5, 6\}$ such that $f(1) + f(2) = f(3)$ is equal to:

If $f(x)=3[x]+\{x+1\}$,where $[x]$ is the greatest integer function of $x$ and $\{x\}$ is the fractional part function of $x$,then $f(-1.32)=$

The range of the function $f(x) = \left[ \frac{1}{\ln(x^2 + e)} \right] + \frac{1}{\sqrt{1 + x^2}}$ is,where $[*]$ denotes the greatest integer function and $e = \lim_{\alpha \to 0} (1 + \alpha)^{1/\alpha}$.

Which of the following pairs of functions are identical?

Vedclass Test Series

Exam Paper Generator

Online Exam Module