Answer is: option3
\( 1.647 < x \leq 2.419 \)Solution:
\( f'(x) = \frac{d}{dx} \left( \int_0^x \cos(t^2 + 2) \, dt \right) = \cos(x^2 + 2) \)
So, \( f(x) \) is increasing when \( f'(x) > 0 \), i.e., when:
\( \cos(x^2 + 2) > 0 \)
Graph the function \( \cos(x^2 + 2) \) on Desmos graphing calculator
\( \cos(x^2 + 2) > 0 \quad \text{when} \quad x \in (1.647, 2.419) \)
Hence option C