JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Definite Integrals

Let

f : R \to R

be a function defined by

f (x) = \frac{4 ^{x}}{4 ^{x} + 2}

and

M = \int_{f (a)}^{f (1 - a)} x sin^{4} (x (1 - x)) d x

N = \int_{f (a)}^{f (1 - a)} sin^{4} (x (1 - x)) d x

;

a \neq = \frac{1}{2}

. If

α M = β N

α, β \in N

, then the least value of

α^{2} + β^{2}

is equal to ____

Answer: 5

Solution

f (x) = \frac{4 ^{x}}{4 ^{x} + 2}

. Check

f (a) + f (1 - a)

f (1 - a) = \frac{4 ^{1 - a}}{4 ^{1 - a} + 2} = \frac{4/ 4 ^{a}}{4/ 4 ^{a} + 2} = \frac{4}{4 + 2 \cdot 4 ^{a}} = \frac{2}{2 + 4 ^{a}}

f (a) + f (1 - a) = \frac{4 ^{a}}{4 ^{a} + 2} + \frac{2}{4 ^{a} + 2} = 1

By the substitution

x \to f (a) + f (1 - a) - x = 1 - x

(King's rule), since

sin^{4} (x (1 - x))

is invariant under

x \to 1 - x

M = \int_{f (a)}^{f (1 - a)} x sin^{4} (x (1 - x)) d x = \int_{f (a)}^{f (1 - a)} (1 - x) sin^{4} (x (1 - x)) d x = N - M

2 M = N

. Given

α M = β N = β (2 M)

, so

α = 2 β

with

α, β \in N

. The least value of

α^{2} + β^{2}

occurs at the smallest positive integers satisfying

α = 2 β

, i.e.

β = 1, α = 2

α^{2} + β^{2} = 4 + 1 = 5

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer