Skip to content
dronacharya.ai
← JEE Advanced 2026 · Paper 2
MathematicsNumerical+4Question 14 of 54

JEE Advanced 2026 · Paper 2, Mathematics question 14

For a real number α\alpha, let [α][\alpha] denote the greatest integer less than or equal to α\alpha. For a finite set SS, let ∣S∣|S| denote the number of elements in the set SS.

Consider the functions f:(−3,3)→(−∞,∞)f:(-3,3)\to(-\infty,\infty) and g:(−3,3)→(−∞,∞)g:(-3,3)\to(-\infty,\infty) defined by
f(x)=[x3]log⁡e(1+sin⁡2(π(x−[x])))f(x)=\left[x^3\right]\log_e\left(1+\sin^2\left(\pi(x-[x])\right)\right)
and
g(x)=x3sin⁡2(πlog⁡e(1+x−[x])).g(x)=x^3\sin^2\left(\pi\log_e(1+x-[x])\right).
Let
A={x∈(−3,3):f is discontinuous at x}A=\{x\in(-3,3):f\text{ is discontinuous at }x\}
and
B={x∈(−3,3):g is discontinuous at x}.B=\{x\in(-3,3):g\text{ is discontinuous at }x\}.
Then the value of ∣A∣+2∣B∣−∣A∩B∣|A|+2|B|-|A\cap B| is ________.

Answer and worked solution

Official answer: 56

ff. sin⁡2(π(x−[x]))=sin⁡2πx\sin^2(\pi(x-[x]))=\sin^2\pi x, so h(x)=ln⁡(1+sin⁡2πx)h(x)=\ln(1+\sin^2\pi x) is continuous and vanishes exactly at integers. f=[x3] hf=[x^3]\,h can only jump where [x3][x^3] jumps, i.e. where x3=kx^3=k is an integer: k=−26,…,26k=-26,\dots,26, which is 53 points in (−3,3)(-3,3). At the five integer points x=0,±1,±2x=0,\pm1,\pm2 we have h=0h=0, so f→0=f(x)f\to0=f(x) from both sides and ff is continuous there. At the other 48 points h≠0h\ne0 and ff jumps. ∣A∣=48|A|=48.

gg. ln⁡(1+{x})\ln(1+\{x\}) is continuous except at integers nn, where it jumps from ln⁡2\ln2 to 00. Since sin⁡2(πln⁡2)≠0\sin^2(\pi\ln2)\ne0, gg jumps at nn unless n3=0n^3=0. So B={−2,−1,1,2}B=\{-2,-1,1,2\}, ∣B∣=4|B|=4.

Overlap. AA contains no integers, so A∩B=∅A\cap B=\varnothing.
∣A∣+2∣B∣−∣A∩B∣=48+8−0=56.|A|+2|B|-|A\cap B|=48+8-0=56.