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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

The value of f (0) , such that f (x) = 1 x^ 2 (1 - ( x)) can be...

The value of f(0)f \left(0\right), such that f(x)=1x2(1−cos⁡(sin⁡x))f \left(x\right) = \dfrac{1}{x^{2}}\left(1 - \cos \left(\sin x\right)\right) can be made continuous at x=0,x = 0, is&

Options

  1. A12\dfrac{1}{2}
  2. B22
  3. C14\dfrac{1}{4}
  4. D44

Correct answer

A. 12\dfrac{1}{2}

Step-by-step solution

lim⁡x→01−cos⁡(sin⁡x)x2=lim⁡x→02sin⁡2(sin⁡x2)x2lim⁡x→02.(sin⁡(sin⁡x2)(sin⁡x2))2.(sin⁡x2x)2=12\lim_{x \to 0} \dfrac{1 - \cos \left(\sin x\right)}{x^{2}} = \lim_{x \to 0} \dfrac{2\sin^{2} \left(\dfrac{\sin x}{2}\right)}{x^{2}}\lim_{x \to 0}2. {\left(\dfrac{\sin \left(\dfrac{\sin x}{2}\right)}{\left(\dfrac{\sin x}{2}\right)}\right)}^{2}. {\left(\dfrac{\sin x}{2x}\right)}^{2} = \dfrac{1}{2}

Thus, for continuity at x=0,x = 0,

f(0)=12f ( 0 ) = \dfrac{1}{2}

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