NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice

If f x = x 2 + x 2 1 + x 2 + x 2 1 + x 2 2 + . … … . . upto ∞, then

Options

Alim x → 0 ⁡ f x does not exist
Bf x is continuous but not differentiable at x = 0
Cf x is discontinuous at x = 0
Df x is differentiable at x = 0

Correct answer

C. f x is discontinuous at x = 0

Step-by-step solution

f 0 = 0 + 0 + 0 + . … … . . = 0 lim x → 0 ⁡ f x = lim x → 0 ⁡ x 2 + x 2 1 + x 2 + x 2 1 + x 2 2 + . … … . . = lim x → 0 ⁡ x 2 1 + 1 1 + x 2 + 1 1 + x 2 2 + . … … . . = lim x → 0 ⁡ x 2 1 1 - 1 1 + x 2 (as 0 < 1 1 + x 2 < 1 ) = lim x → 0 ⁡ 1 + x 2 = 1 So, f 0 ≠ lim x → 0 ⁡ f 0

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