Let the function $f:[1, \infty) \rightarrow \mathbb{R}$ be defined by $$f(t)=\left\{\begin{array}{cc} (-1)^{n+1} 2, & \text { if } t=2 n-1, n \in \mathbb{N}, \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1<t<2 n+1, n \in \mathbb{N} \end{array}\right.$$ 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 ____.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Definite Integrals

Let the function

f : [1, \infty) \to R

be defined by

f (t) = {(- 1)^{n + 1} 2, \frac{( 2 n + 1 - t )}{2} f (2 n - 1) + \frac{( t - ( 2 n - 1 ))}{2} f (2 n + 1), if t = 2 n - 1, n \in N, if 2 n - 1 < t < 2 n + 1, n \in N

Define

g (x) = \int_{1}^{x} f (t) d t, x \in (1, \infty)

. Let

α

denote the number of solutions of the equation

g (x) = 0

in the interval

(1, 8]

and

β = lim_{x \to 1^{+}} \frac{g ( x )}{x - 1}

. Then the value of

α + β

is equal to ____.

Answer: 5

The function

f (t)

is piecewise linear, alternating between

+ 2

and

- 2

at odd integers, with linear interpolation between them.

At

t = 1, 3, 5, 7, 9, \dots

, we have

f (t) = 2, - 2, 2, - 2, \dots

respectively.

Since

g (x) = \int_{1}^{x} f (t) d t

and

g^{'} (x) = f (x)

, the function

g

increases when

f > 0

and decreases when

f < 0

.

The graph of

f

shows that

f

crosses zero at

t = 2, 4, 6, 8, \dots

Since

g (1) = 0

and

g

alternates between increasing and decreasing,

g (x) = 0

x = 1, 3, 5, 7

in the interval

(1, 8]

.

So

α = 3

(solutions at

x = 3, 5, 7

).

For

β

β = x \to 1^{+} lim \frac{g ( x )}{x - 1} = x \to 1^{+} lim \frac{g ( x ) - g ( 1 )}{x - 1} = g^{'} (1^{+}) = f (1^{+}) = 2

Therefore:

α + β = 3 + 2 = 5

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer