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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If y = tan^ - 1 1 1 + x + x^ 2 + tan^ - 1 1 x^ 2 + 3x + 3 + tan^ -...

If y=tan−111+x+x2+tan−11x2+3x+3+tan−11x2+5x+7+....+y = tan^{- 1}\dfrac{1}{1 + x + x^{2}} + tan^{- 1}\dfrac{1}{x^{2} + 3x + 3} + tan^{- 1}\dfrac{1}{x^{2} + 5x + 7} +.... + upto 2n2n terms \left(\forall x \ge 0\right),then, theny \left(0\right) is

Options

  1. Atan−1(n)tan^{- 1} \left(n\right)
  2. Btan−1(2n)tan^{- 1} \left(2n\right)
  3. C2tan−1(n)2tan^{- 1} \left(n\right)
  4. D00

Correct answer

B. tan−1(2n)tan^{- 1} \left(2n\right)

Step-by-step solution

y=tan−111+x+x2+tan−11x2+3x+3+...+2ny = tan^{- 1}\dfrac{1}{1 + x + x^{2}} + tan^{- 1}\dfrac{1}{x^{2} + 3x + 3} +... + 2n terms
=tan−1(x+1)−x1+x(1+x)+tan−1(x+2)−(x+1)1+(x+1)(x+2)+......(2nterms)= tan^{- 1}\dfrac{\left(x + 1\right) - x}{1 + x\left(1 + x\right)} + tan^{- 1}\dfrac{ \left(x + 2\right) - \left(x + 1\right)}{1 + \left(x + 1\right)\left(x + 2\right)} +...... \left(2nterms\right)
=tan−1(x+1)−tan−1x+tan−1(x+2)−tan−1= tan^{- 1}\left(x + 1\right) - tan^{- 1}x + tan^{- 1}\left(x + 2\right) - tan^{- 1}\left(x + 1\right) +... + tan^{- 1} \left(x + 2n\right) - tan^{- 1}\left(x + \left(2n - 1\right)\right)<br/><br />= tan^{- 1} \left(x + 2n\right)−tan−1x- tan^{- 1}x
y\left(0\right) = tan^{- 1} \left(2n\right)
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