Let f be a positive function. If I_1 = int_1-k^k x f [x(1-x) ] dx and I_2 = int_1-k^k f [x(1-x) ] dx, where 2k - 1 > 0. Then, I_1/I_2 is

Let f be a positive function. If I_1 = int_1-k^k x f… — JEE Main 1997 Mathematics PYQ

1997

Let

f

be a positive function.

If

I_1 = \int_{1-k}^{k} x f\left[x(1-x)\right] dx

and

I_2 = \int_{1-k}^{k} f\left[x(1-x)\right] dx

, where

2k - 1 > 0

.

Then,

\dfrac{I_1}{I_2}

A.
2
B.
$k$
C.
$1/2$
D.
1

Show answer and solution

Correct option: C

Solution

Use the property: replacing x → (1-x) in I₁. Since the limits go from (1-k) to k, and (1-k) + k = 1, the substitution x → (1-x) maps the interval to itself.

So I₁ = ∫(1-x)·f[(1-x)x] dx over the same limits.

Adding: 2I₁ = ∫(x + 1-x)·f[x(1-x)] dx = ∫f[x(1-x)] dx = I₂.

Therefore I₁/I₂ = 1/2.

[1997]

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