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1983

JEE Main 1983 Mathematics question on Application of Derivatives

If y=alog⁡x+bx2+xy = a \log x + bx^2 + x has its extremum values at x=−1x = -1 and x=2x = 2, then
  1. A.
    a=2,b=−1a = 2, b = -1
  2. B.
    a=2,b=−12a = 2, b = -\dfrac{1}{2}
  3. C.
    a=−2,b=12a = -2, b = \dfrac{1}{2}
  4. D.
    None of the above
Show answer and solution

Correct option: B

Solution

At extremum, dy/dx = 0. Differentiating: dy/dx = a/x + 2bx + 1 = 0.

At x = -1: -a - 2b + 1 = 0 → a + 2b = 1
At x = 2: a/2 + 4b + 1 = 0 → a + 8b = -2

Subtracting first from second: 6b = -3 → b = -1/2, then a = 1 - 2(-1/2) = 2.

So a = 2, b = -1/2.

[1983]

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