MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Use the quantifiers $\exists$ (there exists) and $\forall$ (for all) to test each statement against the set $A=\{1,2,3,4,5,6\}$.Explanation:Option A: $\exists x \in A$ such that $x+3=9$.For $x=6$, we get $6+3=9$, and $6 \in A$.So Option A is true.Option B: $\exists x \in A$ such that $x+2 < 9$.For $x=1$, we get $1+2=3 < 9$, and $1 \in A$.So Option B is true.Option C: $\forall x \in A,\; x+6 \geq 10$.But for $x=1$, we get $1+6=7$, and $7 \ngeq 10$.So Option C is false.Option D: $\exists x \in A$ such that $x+6 < 10$.For $x=1$, we get $1+6=7 < 10$.So Option D is true.Answer:The statement which is not true is Option C.