Let {x} denote the fractional part of x and f(x)=(cos ⁻¹(1-{x}²) sin ⁻¹(1-{x}))/({x}-{x}³), x ≠ 0. If L and R respectively denotes the left hand limit and the right hand limit of f(x) at x=0, then 32/π²( L²+R²) is equal to ___________.

Solution

Maths sir · 4 steps

Near 0 from the left, {x} is close to 1, not 0, so each side needs its own limit calculation.

1. Right side. For x → 0⁺, {x} = x → 0. The denominator factors as x(1-x)(1+x), and sin⁻¹(1-x) → sin⁻¹ 1 = π/2. That leaves cos⁻¹(1-x²)/x to find.
2. The key estimate. Put θ = cos⁻¹(1-x²). Then 1-cos θ = x², so 2 sin² θ/2 = x² and sin θ/2 = x/√2. For small θ this gives θ ≈ √2 x, and so cos⁻¹(1-x²)/x → √2. That makes R = √2 × π/2 = π/√2.
3. Left side. For x → 0⁻, {x} = 1 + x. Write {x} = 1 - h with h → 0⁺. Then cos⁻¹(1-{x}²) → cos⁻¹ 0 = π/2 and sin⁻¹(1-{x}) = sin⁻¹ h ≈ h. The denominator is (1-h) h (2-h) ≈ 2h, so L = π/2 × h/2h = π/4.

   32/π²(L²+R²) = 32/π²(π²/16+π²/2) = 32/π²×9π²/16

4. So. The value is 2 × 9 = 18.

The common slip: 32. Taking {x} = x on the left side as well makes L = R = π/√2, which gives 32 instead of 18.

The idea to keep. With {x} at an integer, the left side sends {x} to 1 and the right side sends it to 0. Also remember cos⁻¹(1-t²) ≈ √2 t for small t > 0.

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