The range of f(x) = [cos x + sin x] is (Where | vedclass

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Question

(C) We know that the expression $\cos x + \sin x$ can be written as $\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) = \

Vedclass · Accepted Answer

$\{-1, 0, 1\}$

Vedclass · Answer

(C) We know that the expression $\cos x + \sin x$ can be written as $\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) = \sqrt{2} \sin(x + \frac{\pi}{4})$.Since the range of $\sin(x + \frac{\pi}{4})$ is $[-1, 1]$,the range of $\cos x + \sin x$ is $[-\sqrt{2}, \sqrt{2}]$.Given $\sqrt{2} \approx 1.414$,the range of $\cos x + \sin x$ is approximately $[-1.414, 1.414]$.The function $f(x) = [\cos x + \sin x]$ represents the Greatest Integer Function $(G.I.F.)$ of values in the interval $[-1.414, 1.414]$.The integers contained in this interval are $\{-1, 0, 1\}$.Therefore,the range of $f(x)$ is $\{-1, 0, 1\}$.

The range of $f(x) = [\cos x + \sin x]$ is (Where $[.]$ is $G.I.F.$)

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