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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 f(x) = ta n ^ - 1 ( 2 ^ x 1 + 2 ^ 2x + 1 ), then ∑ _ r = 0 ^ 9...

If f(x)=tan−1(2x1+22x+1),f\left(x\right) = ta{n}^{- 1}\left(\dfrac{{2}^{x}}{1 + {2}^{2x + 1}}\right), then ∑r=09f(r)\mathop{∑}_{r = 0}^{9}f\left(r\right) is

Options

  1. Atan⁡−1(1024)\tan^{- 1} \left(1024\right)
  2. Btan⁡−1(10231024)\tan^{- 1} \left(\dfrac{1023}{1024}\right)
  3. Ctan⁡−1(10231025)\tan^{- 1} \left(\dfrac{1023}{1025}\right)
  4. DNone of these

Correct answer

C. tan⁡−1(10231025)\tan^{- 1} \left(\dfrac{1023}{1025}\right)

Step-by-step solution

Let, y=tan−1(2x+1−2x1+2x⋅2x+1)=tan⁡−12(x+1)y = ta{n}^{- 1}\left(\dfrac{{2}^{x + 1} - {2}^{x}}{1 + {2}^{x} ⋅ {2}^{x + 1}}\right) = \tan^{- 1} {2}^{\left(x + 1\right)} - \tan^{- 1} {2}^{x}<br/><br />\mathop{∑}_{r = 0}^{9}f\left(r\right) = \tan^{- 1} \left({2}^{10}\right) - \tan^{- 1} 1 = \tan^{- 1} \left(\dfrac{1023}{1025}\right)
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