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← JEE Advanced 2024 · Paper 2
MathematicsInteger+4Question 13 of 51

JEE Advanced 2024 · Paper 2, Mathematics question 13

Let the function f:[1,∞)→Rf : [1, \infty) \to \mathbb{R} be defined by

f(t)={(−1)n+1 2,if t=2n−1, n∈N,(2n+1−t)2f(2n−1)+(t−(2n−1))2f(2n+1),if 2n−1<t<2n+1, n∈N.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)=∫1xf(t) dtg(x) = \displaystyle\int_1^x f(t)\,dt, x∈(1,∞)x \in (1, \infty). Let α\alpha denote the number of solutions of the equation g(x)=0g(x) = 0 in the interval (1,8](1, 8] and β=lim⁡x→1+g(x)x−1\beta = \displaystyle\lim_{x\to 1+}\frac{g(x)}{x-1}. Then the value of α+β\alpha + \beta is equal to _____.

Answer and worked solution

Official answer: 5

Shape of f. f(2n−1)=(−1)n+1⋅2f(2n-1)=(-1)^{n+1}\cdot2, so f(1)=2f(1)=2, f(3)=−2f(3)=-2, f(5)=2f(5)=2, f(7)=−2f(7)=-2, f(9)=2f(9)=2. Between consecutive odd integers ff is the straight line joining these values.

So ff is a triangle wave with zeros at t=2,4,6,8t=2,4,6,8. Each half-wave (for example [1,2][1,2] or [2,3][2,3]) has area 12⋅1⋅2=1\frac12\cdot1\cdot2=1, with alternating signs.

Values of g. Adding these areas:

xx 2 3 4 5 6 7 8
g(x)g(x) 1 0 −1-1 0 1 0 −1-1

gg is monotonic between consecutive even integers, and g>0g>0 on (1,2](1,2]. So on (1,8](1,8] the zeros are exactly x=3,5,7x=3,5,7, and α=3\alpha=3.

β. β=lim⁡x→1+g(x)−g(1)x−1=g′(1+)=f(1)=2\beta=\lim_{x\to1^+}\dfrac{g(x)-g(1)}{x-1}=g'(1^+)=f(1)=2.

So α+β=5\alpha+\beta=5.

Topics:Fundamental theorem of calculusProperties of definite integralsGraphs of functions
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