If -0bd x-1-x2-b-d x-1-x2- then b-A- tan -11-3B- -2C- -3-2D- 1

The correct option is D 1 tan^ 1b=tan^ 1∞ tan^ 1b 2 nbsp;tan^ 1b= π/2 ⇒ b=1 ...

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Standard XII

Mathematics

Improper Integrals

If ∫0bd x/1+x...

Question

If $\int_{0}^{b} \frac{d x}{1 + x^{2}} = \int_{b}^{\infty} \frac{d x}{1 + x^{2}}$ ,then b=

$t a n^{- 1} (\frac{1}{3})$

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$\frac{\sqrt{3}}{2}$

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$\sqrt{2}$

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Solution

The correct option is D
1

$t a n^{- 1} b = t a n^{- 1} \infty - t a n^{- 1} b$

$2 t a n^{- 1} b = \frac{π}{2} \Rightarrow b = 1$

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If ∫b0dx1+x2=∫∞bdx1+x2,then b=

The correct option is D 1 tan−1b=tan−1∞−tan−1b 2tan−1b=π2⇒b=1

If $\int_{0}^{b} \frac{d x}{1 + x^{2}} = \int_{b}^{\infty} \frac{d x}{1 + x^{2}}$ ,then b=

The correct option is D
1

$t a n^{- 1} b = t a n^{- 1} \infty - t a n^{- 1} b$

$2 t a n^{- 1} b = \frac{π}{2} \Rightarrow b = 1$