Exams › JEE Advanced › Maths
Correct answer: lim(x -> 2n+1) f(x) = 1/2
Analysis of each option: (A) Periodicity: on [0,1): f(x)=x; on [1,2): f(x)=1/2; on [2,3): f(x)=x-2; on [3,4): f(x)=1/2. The pattern on [0,1) is linear (not constant) while on [2,3) it is also linear starting from 0. f(0)=0, f(2)=0, f(4)=0... The pattern repeats with period 2: for x in [2n,2n+1): f(x) = x-2n; for x in [2n+1,2n+2): f(x)=1/2. Yes, f(x+2) = f(x) for all x (shifting n by 1). Option A is TRUE. (B) lim(x->2n+1): Left limit = lim_(x->2n+1⁻) (x-2n) = 1. Right limit = lim_(x->2n+1⁺) 1/2 = 1/2. Left != right, so limit doesn't exist. Option B is FALSE. (C) Even function: f(-1) = f(-1): n=-1, x=-1 falls in [-2,-1) where f(x)=x-(-2)=x+2=-1+2=1. f(1): n=0, x=1 is in [1,2), f(x)=1/2. f(-1)=1 != f(1)=1/2. Not even. Option C is FALSE. (D) lim(x->2n): Left limit = lim_(x->2n⁻) 1/2 = 1/2 (from interval [2n-1,2n), f=1/2). Right limit = lim_(x->2n⁺) (x-2n) = 0. Left != right, limit doesn't exist. Option D is TRUE. So options A and D are correct. Given single-answer format, both A and D are correct. The question may be multiple correct. Among options, A ('periodic with period 2') and D ('limit at 2n doesn't exist') are correct.