$f : N \to A$ and $g : N \to B$ are two onto functions such that $f (n) = [n \sec^{2} θ]$ and $g (n) = [n \cos e c^{2} θ]$ , for some $θ$ and $n \in N$ where N is set of natural numbers, if $\sec^{2} θ$ is irrational, (where [.] denotes the greatest integer function) then which of following is/are TRUE?

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$A \cap B$ is non–empty finite set

f and g are bijective functions

$A \cap B = ϕ$

$A \cup B = N$

answer is A, B, C.

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Detailed Solution

Since $\sec^{2} θ > 1 \Rightarrow [(n + 1) \sec^{2} θ] > [n \sec^{2} θ]$
Hence, f and g are both one–one
Let $k < n \sec^{2} θ < k + 1$ and $k < m \cos e c^{2} θ < k + 1$
$k \cos^{2} θ < n < (k + 1) \cos^{2} θ$ and $k \sin^{2} θ < m (k + 1) \sin^{2} θ$
Adding given k < integer $< k + 1 \Rightarrow A \cap B = ϕ$
Suppose $k \notin A \Rightarrow n \sec^{2} θ < k$ and $(k + 1) < (n + 1) \sec^{2} θ$
$\Rightarrow n < k \cos^{2} θ$ and $n + 1 > (k + 1) \cos^{2} θ$
$\Rightarrow k < (k - n) \cos e c^{2} θ, k + 1 > (k - n) \cos e c^{2} θ$
$[(k - n) \cos e c^{2} θ] = k \in B$