Let I1-1-kkxf - x 1-x -dx-12-1-kkf - x 1-x -dx- where 2k - 1 - 0- Then I1-I2 is

The correct options are C 1/2 D less than 1Using nbsp; ∫_a^bf ( x )dx=∫_a^bf ( a+b x )dx l_1=l_2 l_1 ∴ 2l_1=l_2 hence nbsp; l_1/l_ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Property 1

Let I1=∫1-k...

Question

Let $I_{1} = \int_{1 - k}^{k} x f [x (1 - x)] d x, 1_{2} = \int_{1 - k}^{k} f [x (1 - x)] d x,$ where 2k - 1 > 0. Then $\frac{I_{1}}{I_{2}}$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

less than 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

f

I_{1} = \int_{1 - k}^{k} x f [x (1 - x)] d x

I_{2} = \int_{1 - k}^{k} f [x (1 - x)] d x

2 k - 1 > 0

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

f (x)

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

I_{2} = \int_{1 - k}^{k} f {x (1 - x)} d x

2 k - 1 > 0

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

f

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

I_{2} = \int_{1 - k}^{k} f {x (1 - x)} d x

2 k - 1 > 0

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

I_{1} = \int_{x}^{1} \frac{1}{1 + t^{2}} d t

I_{2} = \int_{1}^{1 / x} \frac{1}{1 + t^{2}} d t

x > 0

f (x)

I_{1} = \int_{0}^{\infty} f (x) d x,

I_{2} = \int_{0}^{\infty} f (3 x - \frac{1}{2 x}) d x,

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

I_{1} & I_{2}

Join BYJU'S Learning Program