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The Value of 1 ∫ 0 Tan − 1 ( 2 X − 1 1 + X − X 2 ) D X , Is(A) 1 (B) 0 (C) −1 (D) π/4 - Mathematics

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Question

The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is

Options

1
0
−1
π/4

MCQ

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Solution

0

\[Let\, I = \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} d x ................(1)\]

\[ = \int_0^1 \tan^{- 1} \frac{2\left( 1 - x \right) - 1}{1 + \left( 1 - x \right) - \left( 1 - x \right)^2} d x\]

\[ = \int_0^1 \tan^{- 1} \frac{1 - 2x}{2 - x - 1 - x^2 + 2x} dx\]

\[ = \int_0^1 \tan^{- 1} \frac{1 - 2x}{1 + x - x^2} dx\]

\[ = - \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx .................(2)\]

\[\text{Adding (i) and (ii)}\]

\[2I = \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx - \int_0^1 \tan^{- 1} \frac{2x - 1}{1 + x - x^2} dx\]

\[ = 0\]

\[Hence\, I = 0\]

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Definite Integrals

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Chapter 20: Definite Integrals - MCQ [Page 120]

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RD Sharma Mathematics [English] Class 12

Chapter 20 Definite Integrals
MCQ | Q 40 | Page 120

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