The range of f(x) = cos 2x - sin 2x contains | vedclass

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

Question

(B) Given $f(x) = \cos 2x - \sin 2x$.We can write this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 2x - \frac{1}{\sqrt{2}} \sin 2x \right)$.Usi

Vedclass · Accepted Answer

$[-1, 1]$

Vedclass · Answer

(B) Given $f(x) = \cos 2x - \sin 2x$.We can write this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos 2x - \frac{1}{\sqrt{2}} \sin 2x ight)$.Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we have $f(x) = \sqrt{2} \cos\left( 2x + \frac{\pi}{4} ight)$.The range of $\cos	heta$ is $[-1, 1]$,so the range of $f(x)$ is $[-\sqrt{2}, \sqrt{2}]$.Since $\sqrt{2} \approx 1.414$,the interval $[-\sqrt{2}, \sqrt{2}]$ is approximately $[-1.414, 1.414]$.Thus,the set $[-1, 1]$ is a subset of $[-\sqrt{2}, \sqrt{2}]$.Therefore,the correct option is $B$.

The range of $f(x) = \cos 2x - \sin 2x$ contains the set

Similar Questions

For $x \in \mathbb{R}$,the range of $3 \cos (4x - 5) + 4$ lies in the interval:

If $A + B + C = \pi$ and $\sin \left( A + \frac{C}{2} \right) = k \sin \frac{C}{2}$,then $\tan \frac{A}{2} \tan \frac{B}{2} = $

The minimum value of $27 \tan^2 \theta + 3 \cot^2 \theta$ is

If $M_1$ and $M_2$ are the maximum values of $\frac{1}{11 \cos 2x + 60 \sin 2x + 69}$ and $3 \cos^2 5x + 4 \sin^2 5x$ respectively,then $\frac{M_1}{M_2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module