If the shortest distance between the lines (x+2)/2=(y+3)/3=(z-5)/4 and (x-3)/1=(y-2)/-3=(z+4)/2 is 38/(3 √5) k, and ∫ 0ᵏ[x²] d x=α-√α, where [x] denotes the greatest integer function, then 6 α³ is equal to _________.

Solution

Maths sir · 4 steps

Get k from the shortest distance, then break the integral wherever x² crosses a whole number.

1. Read off the lines. Line 1 passes through A(-2,-3,5) with direction b⃗₁ = (2,3,4). Line 2 passes through B(3,2,-4) with direction b⃗₂ = (1,-3,2), so AB⃗ = (5,5,-9). The cross product is b⃗₁ × b⃗₂ = (18, 0, -9), with magnitude √405 = 9√5.

   d = (|AB⃗·(b⃗₁ × b⃗₂)|)/(|b⃗₁ × b⃗₂|) = (90 + 0 + 81)/9√5 = 171/9√5 = 19/√5

2. Find k. Set 38k/3√5 = 19/√5. Then 38k = 57, so k = 3/2.
3. Split the integral. On [0, 1.5], [x²] changes value at x = 1 and x = √2. The next break, √3 ≈ 1.73, is past 1.5. So [x²] = 0 on [0,1), 1 on [1,√2) and 2 on [√2, 1.5], and the integral is 1×(√2-1) + 2×(1.5-√2) = 2 - √2.
4. So. Comparing 2 - √2 with α - √α gives α = 2, so 6α³ = 6 × 8 = 48.

The idea to keep. For ∫ [x²] dx, the jumps come at x = √n, not at the integers. List every √n below the upper limit and add up the rectangle areas.

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