Exam

JEE Advanced

Year

2024

Paper

Paper 2

Subject

Mathematics

Chapter

Limits

JEE Advanced 2024 · Paper 2 · Mathematics · Limits

Let the function f:[1, ∞) arrow ℝ be defined by f(t)=\ array cc...

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

Step-by-step solution

\mathrm{f}(\mathrm{t})=\left\{$\begin{array}{ccc}2 & ; & \mathrm{t}=1 \\ 4-2 \mathrm{t} & ; & 1 < \mathrm{t} < 3 \\ -2 & ; & \mathrm{t}=3 \\ -8-2 \mathrm{t} & ; & 3 < \mathrm{t} < 5 \\ 2 & ; & \mathrm{t}=5 \\ 12-2 \mathrm{t} & ; & 5 < \mathrm{t} < 7 \\ -2 & ; & \mathrm{t}=7 \\ -16+2 \mathrm{t} & ; & 7 < \mathrm{t} < 9\end{array}$\right. g(x)=\int_1^x f(t) d t \Rightarrow g^{\prime}(x)=f(x) for $x \in(1,8] g(x)=0 \Rightarrow x=3,5,7 \therefore \alpha=3 \beta=\lim _{x \rightarrow 1^{+}} \frac{g(x)}{x-1}$ Apply L'Hospital's $$\begin{aligned} & =\frac{g^{\prime}\left(1^{+}\right)}{1}=f\left(1^{+}\right) \\ & \beta=2 \\ & \therefore \alpha+\beta=5\end{aligned}$$