JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

For the function f(x) = [x^3] _e(1 + ^2( (x - [x]))), we can simplify the argument of the sine function. Since [x] is an integer, ( (x - [x])) = ( x - [x]) = ( x). Thus, ^2( (x - [x])) = ^2( x). The function f(x) can be rewritten as f(x) = [x^3] _e(1 + ^2( x)). The term _e(1 + ^2( x)) is continuous everywhere. The term [x^3] has jump discontinuities where x^3 is an integer. For x (-3, 3), x^3 (-27, 27). The integer values of x^3 in this interval are k \ -26, -25, , 26\ , which gives 53 points of the form x = k^ 1/3 . At these points, [x^3] is discontinuous. For f(x) to be continuous at x = k^ 1/3 , the continuous factor must be zero: _e(1 + ^2( x)) = 0 ^2( x) = 0 x is an integer. The integers in (-3, 3) are -2, -1, 0, 1, 2. Their cubes are -8, -1, 0, 1, 8, which are 5 values among the 53 points. At these 5 points, f(x) is continuous. At the remaining 53 - 5 = 48 points, f(x) is discontinuous. Thus, |A| = 48. For the function g(x) = x^3 ^2( _e(1 + x - [x])), we can write x - [x] = \ x\ , which is the fractional part of x. The function g(x) = x^3 ^2( _e(1 + \ x\ )) is continuous everywhere except possibly at the integers, where \ x\ is discontinuous. The integers in (-3, 3) are c \ -2, -1, 0, 1, 2\ . Let's check the continuity at x = c: Right-hand limit (x c^+): \ x\ 0, so g(x) c^3 ^2( _e 1) = 0. Left-hand limit (x c^-): \ x\ 1, so g(x) c^3 ^2( _e 2). For g(x) to be continuous at x = c, we must have c^3 ^2( _e 2) = 0. Since _e 2 0.693, _e 2 is not an integer multiple of , meaning ^2( _e 2) 0. Therefore, we must have c^3 = 0 c = 0. So, g(x) is continuous at x = 0 and discontinuous at x \ -2, -1, 1, 2\ . Thus, B = \ -2, -1, 1, 2\ and |B| = 4. Since A contains no integers and B contains only integers, A B = , which means |A B| = 0. Finally, we calculate the required value: |A| + 2|B| - |A B| = 48 + 2(4) - 0 = 48 + 8 = 56. Answer: 56