Let f be a positive function. If I1=∫ limits1-kk x f[x(1-x)] d x and I2=∫ limits1-kk f[x(1-x)] d x, where 2 k-10 . Then, (I1/I2) is

Question

Let f be a positive function. If I1=∫ limits1-kk x f[x(1-x)] d x and I2=∫ limits1-kk f[x(1-x)] d x, where 2 k-1>0 . Then, (I1/I2) is (A) 2 (B) k (C)  (1/2)  (D) 1

Tardigrade Admin · Accepted Answer

Given, I1=∫ limits1-kk x f[x(1-x)] d x ⇒ I1=∫ limits1-kk(1-x) f[(1-x) x] d x ..=∫ limits1-kk f[(1-x)] d x]-∫ limits1-kk x f(1-x)] d x ⇒ I1=I2-I1 ⇒ (I1/I2)=(1/2)

Q. Let $f$ be a positive function.
If $I_{1} = 1 - k \int k x f [x (1 - x)] d x$ and $I_{2} = 1 - k \int k f [x (1 - x)] d x$ , where $2 k - 1 > 0.$
Then, $\frac{I _{1}}{I _{2}}$ is

2

k

$\frac{1}{2}$

1

Given, $I_{1} = 1 - k \int k x f [x (1 - x)] d x$
$\Rightarrow I_{1} = 1 - k \int k (1 - x) f [(1 - x) x] d x$
$= 1 - k \int k f [(1 - x)] d x] - 1 - k \int k x f (1 - x)] d x$
$\Rightarrow I_{1} = I_{2} - I_{1} \Rightarrow \frac{I _{1}}{I _{2}} = \frac{1}{2}$

Q. Let f be a positive function. If I1​=1−k∫k​xf[x(1−x)]dx and I2​=1−k∫k​f[x(1−x)]dx, where 2k−1>0. Then, I2​I1​​ is

2

k

21​

1

Given, I1​=1−k∫k​xf[x(1−x)]dx ⇒I1​=1−k∫k​(1−x)f[(1−x)x]dx =1−k∫k​f[(1−x)]dx]−1−k∫k​xf(1−x)]dx ⇒I1​=I2​−I1​⇒I2​I1​​=21​

Q. Let $f$ be a positive function.
If $I_{1} = 1 - k \int k x f [x (1 - x)] d x$ and $I_{2} = 1 - k \int k f [x (1 - x)] d x$ , where $2 k - 1 > 0.$
Then, $\frac{I _{1}}{I _{2}}$ is

$\frac{1}{2}$

Given, $I_{1} = 1 - k \int k x f [x (1 - x)] d x$
$\Rightarrow I_{1} = 1 - k \int k (1 - x) f [(1 - x) x] d x$
$= 1 - k \int k f [(1 - x)] d x] - 1 - k \int k x f (1 - x)] d x$
$\Rightarrow I_{1} = I_{2} - I_{1} \Rightarrow \frac{I _{1}}{I _{2}} = \frac{1}{2}$