Let [t] denote the greatest integer ≤ t and lim _(x → 0) x[ 4/x ] = A.
Then the function, f(x) = [x²]sin(πx) is discontinuous, when x is equal to :

Solution

Maths sir · 3 steps

The floor part jumps at every √n, but the jump is wiped out wherever sin(π x) is zero.

1. Find A. Write [4/x] = 4/x - {4/x}, where the fractional part lies in [0, 1). Then x[4/x] = 4 - x{4/x}, and the last term is smaller than |x| in size, so it goes to 0. Hence A = 4.
2. Where jumps happen. [x²] jumps only where x² is an integer, that is at x = √n. At such a point the product [x²]sin(π x) still breaks unless sin(π x) = 0 there, which needs x itself to be an integer.

   Option	x	sin(π x)	Verdict
   (a)	√5	not 0	discontinuous: jump from 4 sin(π√5) to 5 sin(π√5)
   (b)	√9 = 3	0	continuous
   (c)	√25 = 5	0	continuous
   (d)	√4 = 2	0	continuous

3. So. Only at x = √(A + 1) = √5 does the jump survive, so the answer is (a).

The idea to keep. For a product like (jumping function) × (smooth function), a jump point stays a discontinuity only if the smooth factor is non-zero there. A zero factor removes the break.

Answer from the NTA final answer key. Solution by PrepWiser.