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2.Relations and Functions
hard
The solution set of $8x \equiv 6(\bmod 14),\,x \in Z$, are
A
$[8] \cup [6]$
B
$[8] \cup [14]$
C
$[6] \cup [13]$
D
$[8] \cup [6] \cup [13]$
Solution
(c) $8x – 6 = 14P,\,(x \in Z)$
==> $x = \frac{1}{8}[14P + 6]$, $(x \in Z)$
==> $x$ = $\frac{1}{4}(7P + 3)$
==> $x = 6, 13, 20, 27, 34, 41, 48,.….$
$\therefore $ Solution set $= {6, 20, 34, 48,…}$ $\cup$ ${13, 27, 41, ….} $= $[6] \cup [13]$,
where $[6], [13]$ are equivalence classes of $6$ and $13$ respectively.
Standard 11
Mathematics
