If f : R → R is a function defined by f(x)= [x - 1] cos ( (2x - 1)/2 )π, where [.] denotes the greatest integer function, then f is :

Solution

Maths sir · 4 steps

The greatest integer part jumps only at integers, and exactly there the cosine factor is zero.

1. Where trouble can come. The cosine factor is continuous everywhere. [x-1] is constant between integers and jumps only when x is an integer, so only the integers need checking.
2. Cosine at integers. At x = n, the angle is (2n-1)π/2, an odd multiple of π/2. So cos(2n-1)π/2 = 0 for every integer n.

   lim(x → n⁻) f(x) = (n-2) × 0 = 0, lim(x → n⁺) f(x) = (n-1) × 0 = 0, f(n) = 0

3. Jump is cancelled. [x-1] goes from n-2 to n-1 at x = n, but it is multiplied by a factor tending to 0. Both one-sided limits and f(n) equal 0, so f is continuous at every integer too.
4. So. f is continuous for every real x, which is option (a).

The idea to keep. When a jumpy factor like [x] multiplies a continuous factor, check the continuous factor at the jump points. If it is zero there, the jump gets killed and the product stays continuous.

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