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JEE Main202117 Mar 2021Evening ShiftMathematicsApplication of DerivativesActual

Consider the function f : R → R defined by f x = 2 - sin 1 x | x | , x ≠ 0 0 , x = 0 . Then f is:

Options

  1. Amonotonic on ( - ∞ , 0 ) ∪ ( 0 , ∞ )
  2. Bnot monotonic on ( - ∞ , 0 ) and ( 0 , ∞ )
  3. Cmonotonic on ( 0 , ∞ ) only
  4. Dmonotonic on ( - ∞ , 0 ) only

Correct answer

B. not monotonic on ( - ∞ , 0 ) and ( 0 , ∞ )

Step-by-step solution

Given f x = 2 - sin 1 x | x | , x ≠ 0                           0 , x = 0 and we know that x =       x , x ≥ 0 - x , x < 0 ⇒ f x = - x 2 - sin 1 x , x < 0                                 0 , x = 0 x 2 - sin 1 x , x > 0 Now, differentiating using product rule, ⇒ f ' x = - 2 - sin 1 x - x - cos 1 x · - 1 x 2 , x &#6

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