NTA Abhyas JEE Main2020MathematicsDifferentiationPractice

If the function f x = cos - 1 ⁡ x 3 2 - 1 - x - x 2 + x 3 (where, ∀ 0 < x < 1 ), then the value of 3 f ' 1 2 is equal to (take 3 = 1.73 )

Correct answer

3.73

Step-by-step solution

f x = cos - 1 ⁡ x x - 1 - x 1 - x 2 = cos - 1 ⁡ x x - 1 - x 2 1 - x 2 = cos - 1 ⁡ x + cos - 1 ⁡ x Now, f ′ x = − 1 1 − x 2 − 1 1 − x ⋅ 1 2 x f ' 1 2 = − 1 1 − 1 4 − 1 1 2 × 1 2 1 2 = − 2 3 − 1 3 f ' 1 2 = - 3 − 2 3 − 1 = 3 . 73

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