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

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(D) Given the function $f(x) = \cos x - \sin x$.We know that any expression of the form $a \cos x + b \sin x$ can be written as $R \cos(x + \alpha)$ o

Vedclass · Accepted Answer

$[-\sqrt{2}, \sqrt{2}]$

Vedclass · Answer

(D) Given the function $f(x) = \cos x - \sin x$.We know that any expression of the form $a \cos x + b \sin x$ can be written as $R \cos(x + \alpha)$ or $R \sin(x + \beta)$,where $R = \sqrt{a^2 + b^2}$.Here,$a = 1$ and $b = -1$.So,$R = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.Thus,$f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x ight) = \sqrt{2} \cos(x + \frac{\pi}{4})$.Since the range of $\cos(	heta)$ is $[-1, 1]$,the range of $\sqrt{2} \cos(x + \frac{\pi}{4})$ is $[-\sqrt{2}, \sqrt{2}]$.Therefore,the correct option is $D$.

The range of $f(x) = \cos x - \sin x$ is

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