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)
(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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