JEE Advanced 2026 · Continuity and Differentiability
Consider the function f:(−2π,2π)→(−∞,∞) defined by f(x)=(∣x∣+∣x−1∣)sinx+[xsinx], where [xsinx] is the greatest integer less than or equal to xsinx. Let α be the total number of points in the interval (−2π,2π) at which f is NOT continuous, and let β be the total number of points in the interval (−2π,2π) at which f is NOT differentiable. Then the value of α+β is ________.
f(x)=(∣x∣+∣x−1∣)sinx+[xsinx],x∈(−2π,2π)To find: α+βPlan: simplify ∣x∣+∣x−1∣;jumps of [xsinx];check kinks∣x∣+∣x−1∣=⎩⎨⎧1−2x,1,2x−1,−2π<x<00≤x<11≤x<2πOnly [xsinx] jumps: h(x)=xsinx∈[0,2π)[h] jumps where h is an integerh(0)=0,h(±2π)=2π≈1.57⇒integers crossed: 0 and 1h=0 only at x=0:[h]=0 both sides⇒no jumph=1 at x=±x1(≈±1.11)⇒jump⇒α=2Non-diff. points⊇{±x1}∪{kinks of ∣x∣,∣x−1∣}x=0:f′−=dxd[(1−2x)sinx]=1,f′+=dxd[sinx]=1⇒f is differentiable at x=0x=1:f′−=cos1,f′+=2sin1+cos12sin1+cos1=cos1⇒not differentiable at x=1β=2(at ±x1)+1(at x=1)=3α+β=2+3=55