Given, x + 2 ≤ x3+3\dfrac{x}{3} + 33x+3 and x is a prime number. The solution set for x is :
∅
{0}
{1}
{0, 1}
Solving the given equation :
⇒x+2≤x3+3⇒x−x3≤3−2⇒3x−x3≤1⇒2x3≤1⇒x≤32⇒x≤1.5\Rightarrow x + 2 \le \dfrac{x}{3} + 3 \\[1em] \Rightarrow x - \dfrac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \dfrac{3x - x}{3} \le 1 \\[1em] \Rightarrow \dfrac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \dfrac{3}{2} \\[1em] \Rightarrow x \le 1.5⇒x+2≤3x+3⇒x−3x≤3−2⇒33x−x≤1⇒32x≤1⇒x≤23⇒x≤1.5
Since, x is a prime number less than 1.5
Solution set is empty.
Hence, Option 1 is the correct option.
