If f(x)=(ex/1+ex), I1=∫ limitsf(-a)f(a) x g(x(1-x)) d x and I2=∫ limitsf(-a)f(a) g(x(1-x)) d x, then find the value of (I2/I1).

Question

If f(x)=(ex/1+ex), I1=∫ limitsf(-a)f(a) x g(x(1-x)) d x and I2=∫ limitsf(-a)f(a) g(x(1-x)) d x, then find the value of (I2/I1). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f(a)+f(-a)=1 I 1=∫ limits f (- a ) f ( a ) xg ( x (1- x )) dx I 1=∫ limits f (- a ) f ( a )(1- x ) g ((1- x ) x ) dx ⇒ I 1+ I 1=∫ limits f (- a ) f ( a ) g ( x (1- x )) dx ⇒ 2 I 1= I 2 ⇒ ( I 2/ I 1)=2

Q. If $f (x) = \frac{e ^{x}}{1 + e ^{x}}, I_{1} = f (- a) \int f (a) xg (x (1 - x)) d x$ and $I_{2} = f (- a) \int f (a) g (x (1 - x)) d x$ , then find the value of $\frac{I _{2}}{I _{1}}$ .

$f (a) + f (- a) = 1$
$I_{1} = f (- a) \int f (a) xg (x (1 - x)) d x$
$I_{1} = f (- a) \int f (a) (1 - x) g ((1 - x) x) d x$
$\Rightarrow I_{1} + I_{1} = f (- a) \int f (a) g (x (1 - x)) d x$
$\Rightarrow 2 I_{1} = I_{2}$
$\Rightarrow \frac{I _{2}}{I _{1}} = 2$

Q. If f(x)=1+exex​,I1​=f(−a)∫f(a)​xg(x(1−x))dx and I2​=f(−a)∫f(a)​g(x(1−x))dx, then find the value of I1​I2​​.

f(a)+f(−a)=1 I1​=f(−a)∫f(a)​xg(x(1−x))dx I1​=f(−a)∫f(a)​(1−x)g((1−x)x)dx ⇒I1​+I1​=f(−a)∫f(a)​g(x(1−x))dx ⇒2I1​=I2​ ⇒I1​I2​​=2

Q. If $f (x) = \frac{e ^{x}}{1 + e ^{x}}, I_{1} = f (- a) \int f (a) xg (x (1 - x)) d x$ and $I_{2} = f (- a) \int f (a) g (x (1 - x)) d x$ , then find the value of $\frac{I _{2}}{I _{1}}$ .

$f (a) + f (- a) = 1$
$I_{1} = f (- a) \int f (a) xg (x (1 - x)) d x$
$I_{1} = f (- a) \int f (a) (1 - x) g ((1 - x) x) d x$
$\Rightarrow I_{1} + I_{1} = f (- a) \int f (a) g (x (1 - x)) d x$
$\Rightarrow 2 I_{1} = I_{2}$
$\Rightarrow \frac{I _{2}}{I _{1}} = 2$