JEE Advanced 2024 Paper 2 · Q13 · Definite Integrals
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 $g(x) = \displaystyle\int_{1}^{x} f(t)\,dt$, $x \in (1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x) = 0$ in the interval $(1, 8]$ and $\beta = \displaystyle\lim_{x \to 1^{+}} \dfrac{g(x)}{x - 1}$. Then the value of $\alpha + \beta$ is equal to ______.
Reveal answer + step-by-step solution
Correct answer:5
Solution
$f$ is piecewise linear, interpolating between values at odd integers: $f(1)=2$, $f(3)=-2$, $f(5)=2$, $f(7)=-2$. On each unit interval the linear pieces give triangles; integrating, $g(x)$ over $[1,3]$ goes from $0$ up to a peak of $2$ at $x=2$ then back to $0$ at $x=3$; over $[3,5]$ it dips symmetrically to $-2$ at $x=4$ then back to $0$ at $x=5$; pattern repeats. So $g(x)=0$ at $x=3,5,7$ in $(1,8]$, giving $\alpha=3$. By L'H\^opital, $\beta = \lim_{x\to 1^{+}} f(x) = f(1) = 2$. Hence $\alpha + \beta = 5$.
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