The set of positive integers less than 50 forming the equivalence class of 6 modulo 9 is given by :

[6]= {6, 15, 24, 33, 42}

The correct answer is Option (2) → [6]= {6, 15, 24, 33, 42}

Using conditions mentioned above,

$x≡a(mod\,n)$

$⇒x≡6(mod\,9)$

$⇒x=6+9k$

for $k=0$,  $6+9(0)=6$

for $k=1$,  $6+9(1)=15$

for $k=2$,  $6+9(2)=24$

for $k=3$,  $6+9(3)=33$

for $k=4$,  $6+9(4)=42$

{6, 15, 24, 33, 42}