Definite Integration
PYP_JEE_ADV_2024_P2
Grade None
Question:
Let the function $f : [1, \infty) \to \mathbb{R}$ be defined by
$$f(t) = \begin{cases} (-1)^{n+1} 2, & \text{if } t = 2n-1, n \in \mathbb{N}, \\ \dfrac{(2n+1-t)}{2} f(2n-1) + \dfrac{(t-(2n-1))}{2} f(2n+1), & \text{if } 2n-1 < t < 2n+1, n \in \mathbb{N}. \end{cases}$$
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 $\beta = \lim\limits_{x \to 1^+} \dfrac{g(x)}{x-1}$. Then the value of $\alpha + \beta$ is equal to ___.
Step-by-Step Solution
Key Concept: Analyzing piecewise continuous functions, integrating them to find roots of the accumulator function, and applying the definition of derivative for the limit.
The function $f(t)$ is piecewise linear between successive odd integers:
- $f(1) = 2$
- $f(3) = -2$
- $f(5) = 2$
- $f(7) = -2$
- $f(9) = 2$
For $t \in [2n-1, 2n+1]$, $f(t)$ is a straight line connecting $(2n-1, f(2n-1))$ and $(2n+1, f(2n+1))$.
Let's evaluate $g(x) = \int_{1}^{x} f(t) dt$:
- For $x \in [1, 3]$: $g(x) = \int_{1}^{x} (4-2t) dt = -(x-1)(x-3)$.
So $g(x) = 0 \implies x = 1$ or $x = 3$.
- For $x \in [3, 5]$: $g(x) = \int_{3}^{x} (2t-8) dt = (x-3)(x-5)$.
So $g(x) = 0 \implies x = 3$ or $x = 5$.
- For $x \in [5, 7]$: $g(x) = -(x-5)(x-7) \implies$ roots at $x = 5, 7$.
- For $x \in [7, 9]$: $g(x) = (x-7)(x-9) \implies$ roots at $x = 7, 9$.
In the interval $(1, 8]$, the roots of $g(x) = 0$ are:
$$x = 3, 5, 7$$
So, the number of solutions is $\alpha = 3$.
To find $\beta$:
$$\beta = \lim_{x \to 1^+} \dfrac{g(x) - g(1)}{x-1} = g'(1) = f(1) = 2$$
Thus, $\alpha + \beta = 3 + 2 = 5$.
Hence, the answer is 5.
Correct Answer: 5