Let [t] denote the greatest integer ≤ t and t denote the fractional part of t. The integral value of α for which the left hand limit of the function
f(x) = [1 + x] + (α ^(2[x] + {x}) + [x] - 1)/(2[x] + { x}) at x = 0 is equal to α - 4/3, is _____________.

Solution

Maths sir · 4 steps

Just left of 0, set x = -h: then [x] = -1, {x} = 1 - h, and every bracket becomes a plain number.

1. Read the brackets. For the left hand limit put x = -h with h → 0⁺. Then [x] = -1 and {x} = x - [x] = 1 - h. Also 1 + x = 1 - h lies in (0, 1), so [1 + x] = 0.
2. Simplify the exponent. The exponent and the denominator are the same expression: 2[x] + {x} = -2 + 1 - h = -1 - h, which tends to -1. So α^(2[x] + {x}) → α⁻¹.

   lim(x → 0⁻) f(x) = 0 + (1/α - 1 - 1)/-1 = 2 - 1/α

3. Set up the equation. Equate the limit to the given value: 2 - 1/α = α - 4/3. Multiply through by 3α to get 6α - 3 = 3α² - 4α, that is 3α² - 10α + 3 = 0.
4. So. The equation factorises as (3α - 1)(α - 3) = 0, so α = 1/3 or α = 3. The integral value is 3. Check: 2 - 1/3 = 5/3 = 3 - 4/3.

The idea to keep. For a one-sided limit with [x] and {x}, fix the side first. Just left of an integer n, [x] = n - 1 and {x} → 1. Then the whole function becomes ordinary algebra.

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