The range of the function y=3 sin left(sqrt{f | vedclass

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

Question

(C) Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$.For the function to be defined,we must have $\frac{\pi^{2}}{16}-x^{2} \g

Vedclass · Accepted Answer

$[0, 3/\sqrt{2}]$

Vedclass · Answer

(C) Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}ight)$.For the function to be defined,we must have $\frac{\pi^{2}}{16}-x^{2} \geq 0$,which implies $x^{2} \leq \frac{\pi^{2}}{16}$,so $x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}ight]$.Let $u = \sqrt{\frac{\pi^{2}}{16}-x^{2}}$. As $x$ varies from $0$ to $\frac{\pi}{4}$,$u$ varies from $\frac{\pi}{4}$ to $0$.Thus,the range of $u$ is $\left[0, \frac{\pi}{4}ight]$.Now,$y = 3 \sin(u)$. Since $u \in \left[0, \frac{\pi}{4}ight]$,$\sin(u)$ varies from $\sin(0) = 0$ to $\sin\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$.Therefore,$y$ varies from $3 	imes 0 = 0$ to $3 	imes \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.Hence,the range of the function is $\left[0, \frac{3}{\sqrt{2}}ight]$.

The range of the function $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$ is

Similar Questions

If $x$ is real,then the range of $\frac{x^2+2x+1}{x^2+2x+7}$ is

If $[x]^2-5[x]+6=0$,where $[.]$ denotes the greatest integer function,then

The domain of the real valued function $f(x) = \sqrt{\cos (\sin x)} + \cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$ is

The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$,where $x \neq \frac{n\pi}{2}, n \in \mathbb{Z}$ and $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$,is

If $f: R \rightarrow R$ is defined by $f(x)=[2x]-2[x]$ for $x \in R$,where $[x]$ is the greatest integer not exceeding $x$,then the range of $f$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module