If the function $\int_0^x f(t)dt - 5$ as$[x] \to 1$, where $f$ is continuous then the number of integers in the range of $p$ so that the equation $2x + \int_0^x f(t)dt = p$ has roots of opposite sign in $(-1, 1)$.
Step-by-Step Solution
Key Concept: Apply the Intermediate Value Theorem to F(x) = 2x + ∫₀ˣ f(t)dt - p by evaluating it at boundary and interior points; roots of opposite sign in (-1,1) require F to change sign across x=0, necessitating F(-1)·F(0) < 0 AND F(0)·F(1) < 0 simultaneously.
Let $F(x) = 2x + \int_0^x f(t)dt - p$ defined on $[-1,1]$. We have $F(0) = -p$ and $F(1) = 7 - p$, and $F(-1) = 3 - p$. For a root in $(-1,1)$, we need $F(-1)F(0) < 0$ and $F(0)F(1) < 0$, which gives $p \in (0,3)$.
Correct Answer: 2