MCQ
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 $R-\{0\}$
  • $S=R-\{0\}$

Answer

Correct option: D.
$S=R-\{0\}$
d
(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

$x=0$

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

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

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