Let α ∈ R be such that the function f(x) = { (cos ⁻¹(1 - { x} ²)sin ⁻¹(1 - { x} ))/({ x} - { x} ³),, x ≠ 0; α ,, x = 0 is continuous at x = 0, where x = x - [ x ] is the greatest integer less than or equal to x. Then :

Solution

Maths sir · 4 steps

Near 0 the fractional part jumps from 1 down to 0, so check the two sides separately.

1. Two sides differ. As x → 0⁺, {x} = x → 0. As x → 0⁻, {x} = x + 1 → 1. Write t = {x} and factor the denominator as t - t³ = t(1-t)(1+t).

   f = (cos⁻¹(1-t²) sin⁻¹(1-t))/t(1-t)(1+t)

2. Right-hand limit. Here t → 0⁺. If cos θ = 1 - t², then θ²/2 ≈ t², so cos⁻¹(1-t²) ≈ √2 t. Also sin⁻¹(1-t) → π/2, and the denominator is about t. So the RHL is √2 × π/2 = π/√2.
3. Left-hand limit. Here t → 1⁻. Then cos⁻¹(1-t²) → cos⁻¹ 0 = π/2, and sin⁻¹(1-t) ≈ 1-t. The denominator is about 2(1-t). So the LHL is π/2 × 1/2 = π/4.
4. So. The LHL is π/4 and the RHL is π/√2. They are not equal, so the limit at 0 does not exist and no value of α makes f continuous. The answer is (a).

The idea to keep. Whenever {x} or [x] appears at an integer, find LHL and RHL separately. {x} tends to 1 from the left and to 0 from the right.

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