Let S={x ∈ R: cos (x)+cos (√{2} x)<2} , then

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Trigonometrical Equations

hard

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

A

$S=\emptyset$

B

$S$ is a non-empty finite set

C

$S$ is an infinite proper subset of $R-\{0\}$

D

$S=R-\{0\}$

(KVPY-2018)

Solution

(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

$x=0$

$\therefore \quad \cos x+\cos \sqrt{2} x=2 \text { at } x=0$

$\text { Hence, } \quad S=R-\{0\}$

Standard 11

Mathematics

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