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JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

If the function ff given by f(x)=x3−3(a−2)x2+3ax+7f(x) = x^3 - 3(a-2)x^2 + 3ax + 7, for some a∈Ra \in R is increasing in (0,1](0, 1] and decreasing in [1,5)[1, 5), then a root of the equation, (x−1)−14(x−1)2=0\frac{(x-1)^{-14}}{(x-1)^2} = 0 (x≠1)(x \neq 1) is
  1. A.
    -7
  2. B.
    6
  3. C.
    7
  4. D.
    5
Show answer and solution

Correct option: C

Solution

f'(x) = 3x² - 6(a-2)x + 3a. Since f increases on (0,1] and decreases on [1,5), x=1 is a local maximum, so f'(1) = 0.

f'(1) = 3 - 6(a-2) + 3a = 3 - 6a + 12 + 3a = 15 - 3a = 0 → a = 5.

Now f(x) = x³ - 9x² + 15x + 7, and the equation to solve is f(x-7)/(x-1)² = 0 (interpreting the given expression as [f(x) - 14]/(x-1)² = 0, i.e. f(x) = 14).

With a = 5: f(x) = 14 gives x³ - 9x² + 15x + 7 = 14, i.e. x³ - 9x² + 15x - 7 = 0. This factors as (x-1)²(x-7) = 0. Since x ≠ 1, the root is x = 7.

[JEE Main 2019 · 12 Jan · Shift II]

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