If f(x)=ex1+ex,l1=∫f(−a)f(a) xg{x(1−x)}dx .Then the value of I2/I1, is

A
-1
B
1/2
C
2
D
1

Solution:

We have, $f (x) = \frac{e^{x}}{1 + e^{x}}$

$\begin{array}{l} f (- x) = \frac{e^{- x}}{1 + e^{- x}} = \frac{1}{1 + e^{x}} \\ ∴ f (x) + f (- x) = \frac{e^{x}}{1 + e^{x}} + \frac{1}{1 + e^{x}} = 1 \\ \Rightarrow f (a) + f (- a) = 1 \end{array}$

Now,

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

Using $\int_{a}^{b} f (a + b - x) = \int_{a}^{b} f (x) and f (- a) + f (a) = 1$ , we get

$\begin{array}{l} I_{1} = \int_{f (- a)}^{f (a)} (1 - x) g {(1 - x) (1 - (1 - x))} d x \\ \Rightarrow I_{1} = \int_{f (- a)}^{f (a)} (1 - x) g {x (1 - x)} d x \end{array}$

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

Adding (i) and (ii), we get

$\begin{aligned} 2 I_{1} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x \\ \Rightarrow 2 I_{1} = I_{2} \Rightarrow \frac{I_{2}}{I_{1}} = 2 \end{aligned}$

If $f (x) = \frac{e^{x}}{1 + e^{x}}, l_{1} = \int_{f (- a)}^{f (a)} x g {x (1 - x)} d x$ .Then the value of $I_{2} / I_{1}$ , is

Solution:

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