Let the function f:[1, \infty) \rightarrow  be defined by f(t)={\\begin{array}{c}(-1)^{n+1\end{array} 2{,} , \text{ if } t=2 n-1, n \in {,} ; \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1){,} , \text{ if } 2 n-1< t< 2 n+1, n \in . }. 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 \rightarrow 1^{+}} \frac{g(x)}{x-1} . Then the value of \alpha+\beta is equal to _______ . [JEE Adv 2024 P2]

Application of Integration

Section: Mathematics

Question : 3 of 28

Marks: +1, -0

Let the function f:[1,∞)⟶  be defined by

f(t)={

(−1)n+12,

‌ if ‌t=2n−1

n∈ ,

‌

(2n+1−t)

f(2n−1)+‌

(t−(2n−1))

f(2n+1),

‌ if ‌2n−1<t<2n+1

n∈ .

Define g(x)=

∫

f(t)‌dt,x∈(1,∞) . Let α denote the number of solutions of the equation g(x)=0 in the interval (1,8] and β=

lim

x⟶1+

‌

g(x)

x−1

. Then the value of α+β is equal to _______ .

[JEE Adv 2024 P2]

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