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BITSAT 2017
Exam
BITSAT
Year
2017
Subject
Mathematics
Chapter
Continuity and Differentiability

BITSAT 2017 · Mathematics · Continuity and Differentiability

The value of f ( 0 ) , so that the function f(x) = 2x - ^ -1 x 2x +...

The value of f(0)f ( 0 ), so that the function f(x)=2x−sin⁡−1x2x+tan⁡−1xf\left(x\right) = \dfrac{2x - \sin^{-1}x}{2x + \tan^{-1}x} is continuous at each point in its domain, is

Options

  1. A13\dfrac{1}{3}
  2. B−13- \dfrac{1}{3}
  3. C23\dfrac{2}{3}
  4. D−23- \dfrac{2}{3}

Correct answer

A. 13\dfrac{1}{3}

Step-by-step solution

Since, x,sin⁡−1x,tan⁡−1xx, \sin^{- 1}x, \tan^{- 1}x are continuous functions, so the
function ff is clearly continuous at each point of its domain except possibly at x=0x = 0.

So, for ff to be continuous at x=0x = 0

f(0)=lim⁡x→0f(x)=lim⁡x→02−sin⁡−1xx2+tan⁡−1xx=13f\left(0\right) = \lim_{x \to 0}f\left(x\right) = \lim_{x \to 0}\dfrac{2 - \dfrac{\sin^{- 1}x}{x}}{2 + \dfrac{\tan^{- 1}x}{x}} = \dfrac{1}{3}

\left[∵ \lim_{x \to 0}\dfrac{\sin^{- 1}x}{x} = 1, \lim_{x \to 0}\dfrac{\tan^{- 1}x}{x} = 1\right]

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