For p, q ∈ R, consider the real valued function f(x)=(x-p)²-q, x ∈ R and q>0. Let a₁, a_(2′) a₃ and a₄ be in an arithmetic progression with mean p and positive common difference. If |f(aᵢ)|=500 for all i=1,2,3,4, then the absolute difference between the roots of f(x)=0 is ___________.

Solution

Maths sir · 4 steps

Write the four terms symmetrically around p, and f takes only two values, one +500 and one -500.

1. Symmetric terms. The mean of the AP is p, so take the terms as p-3d, p-d, p+d, p+3d with d>0. The common difference is 2d, which is positive.
2. Only two values. Since f(aᵢ) = (aᵢ-p)² - q, the outer terms give 9d² - q and the inner terms give d² - q. These two values differ because d>0, so both cannot be 500 and both cannot be -500. The larger one is +500 and the smaller one is -500.

   9d² - q = 500, d² - q = -500

3. Solve. Subtracting gives 8d² = 1000, so d² = 125. Then q = d² + 500 = 625.
4. So. The roots of (x-p)² = q are p ± √q = p ± 25. Their absolute difference is 2√q = 50.

The common slip: 25. Stopping at √q gives 25, which is the distance from p to one root. The two roots p ± 25 are 50 apart.

The idea to keep. For an AP with a known mean, write the terms symmetrically about that mean. For (x-p)² - q, the gap between the roots is always 2√q.

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