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JEE Main 2019 · 12 April · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

Let f(x)=5−∣x−2∣f(x) = 5 - |x - 2| and g(x)=∣x+1∣g(x) = |x + 1|, x∈Rx \in R. If f(x)f(x) attains maximum value at α\alpha and g(x)g(x) attains minimum value of β\beta, then lim⁡x→−αβ(x−1)(x2−5x+6)x2−6x+8\lim_{x \to -\alpha\beta} \frac{(x-1)(x^2 - 5x + 6)}{x^2 - 6x + 8} is equal to
  1. A.
    1/2
  2. B.
    -3/2
  3. C.
    -1/2
  4. D.
    3/2
Show answer and solution

Correct option: A

Solution

f(x) = 5 - |x-2| is maximized when |x-2| = 0, so α = 2. g(x) = |x+1| is minimized when |x+1| = 0, so β = -1.

Thus -αβ = -(2)(-1) = 2.

The limit becomes lim(x→2) of (x-1)(x²-5x+6)/(x²-6x+8).

Factor: x²-5x+6 = (x-2)(x-3) and x²-6x+8 = (x-2)(x-4).

Cancel (x-2): lim(x→2) of (x-1)(x-3)/(x-4) = (1)(-1)/(-2) = 1/2.

[JEE Main 2019 · 12 April · Shift II]

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