Let the function f(x) = x/3 + 3/x + 3, x ≠ 0 be strictly increasing in (-∞, α₁) ∪ (α₂, ∞) and strictly decreasing in (α₃, α₄) ∪ (α₄, α₅). Then Σᵢ₌₁⁵ αᵢ² is equal to

Solution

Maths sir · 4 steps

Find where f'(x) changes sign, and remember that x = 0 splits the decreasing part into two pieces.

1. Differentiate. The constant 3 drops out, and 3/x gives -3/x².

   f'(x) = 1/3 - 3/x² = (x² - 9)/3x²

2. Sign of the derivative. The denominator 3x² is positive for every x ≠ 0, so the sign of f'(x) is the sign of x² - 9. It is positive when |x| > 3 and negative when 0 < |x| < 3.
3. Read off the intervals. f is strictly increasing on (-∞, -3) ∪ (3, ∞), so α₁ = -3 and α₂ = 3. It is strictly decreasing on (-3, 0) ∪ (0, 3), so α₃ = -3, α₄ = 0 and α₅ = 3. The point x = 0 is excluded because f is not defined there.
4. So. Σ αᵢ² = 9 + 9 + 9 + 0 + 9 = 36, which is option (c).

The idea to keep. Find the sign of f'(x) to get where f increases or decreases. Any point where f is undefined, like x = 0 here, must also be cut out of the intervals.

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