StreakPeaked· Practice

ExamsJEE MainMaths

For the function y = a²/x + b²/(a - x), where a > 0 and b > 0, the least value of y for x in the interval (0, a) is:

  1. a + b
  2. 1/(a+b)
  3. (1/a)(a+b)²
  4. (1/a²)(a+b)

Correct answer: (1/a)(a+b)²

Solution

The expression for y can be minimized using calculus or the method of Lagrange multipliers, leading to the conclusion that the least value occurs at the critical points derived from the function's derivative. The correct option, (1/a)(a+b)², represents the minimum value of y when evaluated at the optimal point within the given interval.

Related JEE Main Maths questions

⚔️ Practice JEE Main Maths free + battle 1v1 →