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 ...

The correct numerical answer is 5. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Mathematics Question 13 Solution

From the definition of $f(t)$ : $f(1) = 2, f(3) = -2, f(5) = 2, f(7) = -2, f(9) = 2$ . Between these odd integers, $f(t)$ is linear.
For $t \in [2n-1, 2n+1]$ , $g(x) = \int_1^x f(t) dt$ represents the net area.
In $[1, 3]$ , $f(t)$ goes from 2 to -2 linearly, crossing zero at $t=2$ . $g(3) = \int_1^3 f(t) dt = \text{Area}(1,2) - \text{Area}(2,3) = 1 - 1 = 0$ .
In $[3, 5]$ , $f(t)$ goes from -2 to 2 linearly, crossing zero at $t=4$ . $g(5) = g(3) + \int_3^5 f(t) dt = 0 + (-1 + 1) = 0$ .
Similarly, $g(7) = 0$ and $g(9) = 0$ .
The solutions for $g(x) = 0$ in $(1, 8]$ are $x=3, 5, 7$ . No other roots exist in between because $g(x)$ is strictly monotonic between the midpoints (even integers) and the odd integers. Thus, $\alpha = 3$ .
Calculating $\beta = \lim_{x \to 1^+} \frac{g(x)}{x-1}$ . By L'Hopital's rule, $\beta = \lim_{x \to 1^+} g'(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$ .
The value of $\alpha + \beta = 3 + 2 = 5$ .