Let f(x) = | a, - 1, 0; ax, a, - 1; ax², ax, a |, a ∈ R. Then the sum of the squares of all the values of a, for which 2f'(10) - f'(5) + 100 = 0, is

Solution

Maths sir · 4 steps

Open the determinant first. It becomes a simple quadratic in x, and differentiating it is then easy.

1. Expand along row 1. The third entry of row 1 is 0, so only two cofactors survive: f(x) = a(a² + ax) + 1 · (a²x + ax²).

   f(x) = a³ + 2a²x + ax² ⇒ f'(x) = 2a² + 2ax

2. Use the condition. f'(10) = 2a² + 20a and f'(5) = 2a² + 10a. So 2(2a² + 20a) - (2a² + 10a) + 100 = 0, which gives 2a² + 30a + 100 = 0, that is, a² + 15a + 50 = 0.
3. Roots. (a + 5)(a + 10) = 0, so a = -5 or a = -10. Vieta gives the same check: α² + β² = (α + β)² - 2αβ = 225 - 100.
4. So. The sum of the squares is 25 + 100 = 125, which is option (c).

The idea to keep. When a determinant has variable entries and the question asks about its derivative, expand it first. Pick the row or column with a zero to save work. A sum of squares of roots is (α+β)² - 2αβ.

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