Q61 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $A = \{1, 2, 3, 4, 5\}$, then which of the following statements is not true?
- A$\exists x \in A$, such that $x + 4 \leq 9$
- B$\exists x \in A$, such that $12 - x = 7$
- C$\forall x \in A, x + 3 \geq 4$
- Correct answerD$\exists x \in A$, such that $x + 2 < 3$
Solution
Given the set $A = \{1, 2, 3, 4, 5\}$.
Checking the first statement: $\exists x \in A$, such that $x + 4 \leq 9$. For $x = 1$, $1 + 4 = 5 \leq 9$. This statement is true.
Checking the second statement: $\exists x \in A$, such that $12 - x = 7$. Solving for $x$ gives $x = 5$. Since $5 \in A$, this statement is true.
Checking the third statement: $\forall x \in A, x + 3 \geq 4$. The minimum value of $x$ in $A$ is $1$. For $x = 1$, $1 + 3 = 4 \geq 4$. For all other values in $A$, $x + 3 > 4$. This statement is true.
Checking the fourth statement: $\exists x \in A$, such that $x + 2 < 3$. Solving for $x$ gives $x < 1$. There is no element in $A$ that is strictly less than $1$. This statement is false.
Therefore, the fourth statement is not true.
Answer: $\exists x \in A$, such that $x + 2 < 3$
Checking the first statement: $\exists x \in A$, such that $x + 4 \leq 9$. For $x = 1$, $1 + 4 = 5 \leq 9$. This statement is true.
Checking the second statement: $\exists x \in A$, such that $12 - x = 7$. Solving for $x$ gives $x = 5$. Since $5 \in A$, this statement is true.
Checking the third statement: $\forall x \in A, x + 3 \geq 4$. The minimum value of $x$ in $A$ is $1$. For $x = 1$, $1 + 3 = 4 \geq 4$. For all other values in $A$, $x + 3 > 4$. This statement is true.
Checking the fourth statement: $\exists x \in A$, such that $x + 2 < 3$. Solving for $x$ gives $x < 1$. There is no element in $A$ that is strictly less than $1$. This statement is false.
Therefore, the fourth statement is not true.
Answer: $\exists x \in A$, such that $x + 2 < 3$
Hence, the correct answer is option: 'D'.