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JEE Advanced 2026 · Continuity and Differentiability

Consider the function f:(−π2,π2)→(−∞,∞)f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to (-\infty, \infty) defined by f(x)=(∣x∣+∣x−1∣)sin⁡x+[xsin⁡x]f(x) = (|x| + |x - 1|) \sin x + [x \sin x], where [xsin⁡x][x \sin x] is the greatest integer less than or equal to xsin⁡xx \sin x. Let α\alpha be the total number of points in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) at which ff is NOT continuous, and let β\beta be the total number of points in the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) at which ff is NOT differentiable. Then the value of α+β\alpha + \beta is ________.

Answer: 55

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f(x)=(∣x∣+∣x−1∣)sin⁡x+[ xsin⁡x ],x∈(−π2,π2)f(x) = (|x| + |x-1|)\sin x + [\,x\sin x\,],\quad x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) To find: α+β\text{To find: } \alpha + \beta Plan: simplify ∣x∣+∣x−1∣; jumps of [xsin⁡x]; check kinks\text{Plan: simplify } |x|+|x-1|;\ \text{jumps of } [x\sin x];\ \text{check kinks} ∣x∣+∣x−1∣={1−2x,−π2<x<01,0≤x<12x−1,1≤x<π2|x|+|x-1| = \begin{cases} 1-2x, & -\tfrac{\pi}{2}<x<0\\ 1, & 0\le x<1\\ 2x-1, & 1\le x<\tfrac{\pi}{2}\end{cases} Only [ xsin⁡x ] jumps: h(x)=xsin⁡x∈[0,π2)\text{Only } [\,x\sin x\,] \text{ jumps: } h(x)=x\sin x \in \left[0,\dfrac{\pi}{2}\right) [h] jumps where h is an integer[h] \text{ jumps where } h \text{ is an integer} h(0)=0,h(±π2)=π2≈1.57h(0)=0,\qquad h\left(\pm\tfrac{\pi}{2}\right)=\tfrac{\pi}{2}\approx 1.57 ⇒integers crossed: 0 and 1\Rightarrow \text{integers crossed: } 0 \text{ and } 1 h=0 only at x=0: [h]=0 both sides⇒no jumph=0 \text{ only at } x=0:\ [h]=0 \text{ both sides} \Rightarrow \text{no jump} h=1 at x=±x1 (≈±1.11)⇒jump⇒α=2h=1 \text{ at } x=\pm x_1\ (\approx \pm 1.11) \Rightarrow \text{jump} \Rightarrow \alpha = 2 Non-diff. points⊇{±x1}∪{kinks of ∣x∣, ∣x−1∣}\text{Non-diff. points} \supseteq \{\pm x_1\} \cup \{\text{kinks of } |x|,\ |x-1|\} x=0:f′−=ddx[(1−2x)sin⁡x]=1,f′+=ddx[sin⁡x]=1x=0:\quad f'^- = \dfrac{d}{dx}[(1-2x)\sin x] = 1,\quad f'^+ = \dfrac{d}{dx}[\sin x] = 1 ⇒f is differentiable at x=0\Rightarrow f \text{ is differentiable at } x=0 x=1:f′−=cos⁡1,f′+=2sin⁡1+cos⁡1x=1:\quad f'^- = \cos 1,\quad f'^+ = 2\sin 1 + \cos 1 2sin⁡1+cos⁡1≠cos⁡1⇒not differentiable at x=12\sin 1 + \cos 1 \neq \cos 1 \Rightarrow \text{not differentiable at } x=1 β=2 (at ±x1)+1 (at x=1)=3\beta = 2\ (\text{at } \pm x_1) + 1\ (\text{at } x=1) = 3 α+β=2+3=5\alpha + \beta = 2 + 3 = 5 5\boxed{5}

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