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JEE Main202224 Jun 2022Morning ShiftMathematicsApplication of DerivativesActual

The sum of absolute maximum and absolute minimum values of the function f x = 2 x 2 + 3 x - 2 + sin x cos x in the interval 0 , 1 is

Options

  1. A3 + sin 1 cos 2 1 2 2
  2. B3 + 1 2 1 + 2 cos 1 sin 1
  3. C5 + 1 2 sin 1 + sin 2
  4. D2 + sin 1 2 cos 1 2

Correct answer

B. 3 + 1 2 1 + 2 cos 1 sin 1

Step-by-step solution

Given f x = 2 x 2 + 3 x - 2 + sin x cos x ⇒ f x = 2 x - 1 x + 2 + sin x cos x Now, f ' x = 4 x + 3 + cos 2 x 4 , 1 2 < x < 1 - 4 x + 3 + cos 2 x 4 , 0 ≤ x < 1 2 For 0 ≤ x < 1 2 ⇒ f ' x < 0 For 1 2 < x ≤ 1 ⇒ f ' x > 0 So f x has minima at x = 1 2 and maxima at x = 1 Hence, f 1 2 + f 1 = 3 + 1 2 1 + 2 cos 1 sin 1

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