JEE Advanced 2024 Paper 2Question 1
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 $\beta=\lim_{x\to1^+}\frac{g(x)}{x-1}$. Then the value of $\alpha+\beta$ is equal to _____.
Definite Integration
Answer$5$
Solution
Concept. $f$ takes the values $2,-2,2,-2,\dots$ at $1,3,5,7,\dots$ and is linear in between. So $g(x)$ is a signed area of triangles, which cancel over each $[2n-1,2n+1]$. Formulas used. (1) $g(x)=\int_1^xf(t)\,dt$ = (area above the $t$-axis) $-$ (area below)(2) $\lim_{x\to1^+}\dfrac{g(x)}{x-1}=g'(1^+)=f(1)$ ($f$ continuous)Step 1 - Graph of $f$. $f(1)=2$, $f(3)=-2$, $f(5)=2$, $f(7)=-2$, $f(9)=2$, joined by straight lines. Zeros at $t=2,4,6,8$.Step 2 - Values of $g$. On $[1,3]$ the triangle above (area $1$) cancels the triangle below (area $1$), so $g(3)=0$. In the same way $g(5)=g(7)=0$. Step 3 - Sign between. On $(1,3)$, $g>0$ (it rises to $1$ at $t=2$, then falls to $0$). On $(3,5)$, $g<0$; on $(5,7)$, $g>0$; on $(7,8]$, $g<0$ (with $g(8)=-1$). So the zeros in $(1,8]$ are $3,5,7$: $\alpha=3$. Step 4 - $\beta$. By (2), $\beta=f(1)=2$. Step 5 - Value. $\alpha+\beta=5$. Answer: 5