Q35 / 91 · KVPY 2018 (SB/SX) · Mathematics

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

A
$S = ϕ$
B
$S$ is a non-empty finite set
C
$S$ is an infinite proper subset of $R - \{0\}$
Correct answer
D
$S = R - \{0\}$

Solution

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 $x = 0$

$∴ \cos x + \cos \sqrt{2} x = 2$ at $x = 0$

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

Hence, the correct answer is option: 'D'.

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