Let S={x in R: cos (x)+cos (sqrt{2} x)

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Question

(d)

We have,

$S=\{x \in R: \cos x+\cos \sqrt{2} x<2\}$

Maximum value of $\cos x$ and $\cos \sqrt{2} x$ is 1 at

$x=0$

$	herefore \qu

Vedclass · Accepted Answer

$S=R-\{0\}$

Vedclass · Answer

(d)

We have,

$S=\{x \in R: \cos x+\cos \sqrt{2} x<2\}$

Maximum value of $\cos x$ and $\cos \sqrt{2} x$ is 1 at

$x=0$

$	herefore \quad \cos x+\cos \sqrt{2} x=2 	ext { at } x=0$

$	ext { Hence, } \quad S=R-\{0\}$

Let $S=\{x \in R: \cos (x)+\cos (\sqrt{2} x)<2\}$, then

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