Question

Let the function \( f : [1, \infty) \to \mathbb{R} \) be defined by:
\[ f(t) = \begin{cases} (-1)^{n+1}2, & \text{if } t = 2n-1, \, n \in \mathbb{N}, \\ \frac{(2n+1-t)}{2}f(2n-1) + \frac{(t-(2n-1))}{2}f(2n+1), & \text{if } 2n-1 < t < 2n+1, \, n \in \mathbb{N}. \end{cases} \]
Define \( g(x) = \int_1^x f(t) \, dt, \, x \in (1, \infty). \) Let \( \alpha \) denote the number of solutions of the equation \( g(x) = 0 \) in the interval \( (1, 8] \) and \( \beta = \lim_{x \to 1^+} \frac{g(x)}{x-1}. \) Then the value of \( \alpha + \beta \) is .........

Hint:

Visualize piecewise functions graphically to calculate integrals and identify roots.

Answer 1

From the piecewise definition of \( f(t) \): \[ f(t) = \begin{cases} 2, & t = 1, \\ 4 - 2t, & 1<t<3, \\ -2, & t = 3, \\ -8 - 2t, & 3<t<5, \, \text{and so on.} \end{cases} \] The integral \( g(x) \): \[ g(x) = \int_1^x f(t) \, dt = 0 \quad \text{at } x = 3, 5, 7. \] Thus, \( \alpha = 3 \). For \( \beta \): \[ \lim_{x \to 1^+} \frac{g(x)}{x-1} = \lim_{x \to 1^+} f(1) = 2 \quad \Rightarrow \quad \beta = 2. \] Therefore: \[ \alpha + \beta = 3 + 2 = 5. \]