Fx-ex-1-ex- I1-f-afa x gx1-x d x- I2-f-afa gx1-x d x- then l2-l1-

The correct option is B 2 2I_1=I_2⇒I_2/I_1=2 ...

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Standard XII

Mathematics

Vn Method

fx=ex/1+ex, I...

Question

$f (x) = \frac{e^{x}}{1 + e^{x}}, I_{1} = f (a) \int f (- a) x g (x (1 - x)) d x, I_{2} = f (a) \int f (- a) g (x (1 - x)) d x, t h e n \frac{I_{2}}{I_{1}} =$

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Solution

The correct option is A
2

$2 I_{1} = I_{2} \Rightarrow \frac{I_{2}}{I_{1}} = 2$

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Similar questions

Q. If

f (x) = \frac{e^{x}}{1 + e^{x}}

I_{1} = \int_{f (- a)}^{f (a) x g {x (1 - x)}} d x

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\frac{I_{2}}{I_{1}}

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Q. If

f (x) = \frac{e^{x}}{1 + e^{x}}

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f (x) = \frac{e^{x}}{1 + e^{x}}, I_{1} = f (a) \int f (- a) x g (x (1 - x)) d x

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f(x)=ex1+ex,I1=f(a)∫f(−a)xg(x(1−x))dx,I2=f(a)∫f(−a)g(x(1−x))dx,thenI2I1=

The correct option is A 2 2I1=I2⇒I2I1=2

$f (x) = \frac{e^{x}}{1 + e^{x}}, I_{1} = f (a) \int f (- a) x g (x (1 - x)) d x, I_{2} = f (a) \int f (- a) g (x (1 - x)) d x, t h e n \frac{I_{2}}{I_{1}} =$

The correct option is A
2

$2 I_{1} = I_{2} \Rightarrow \frac{I_{2}}{I_{1}} = 2$