Let f : R R be a function defined by f ( x ) = 4 x 4 x + 2 and M = f ( a ) f ( 1 - a ) x sin 4 ( x ( 1 - x ) ) d x , N = f ( a ) f ( 1 - a ) sin 4 ( x ( 1 - x ) ) d x , a ? 1 2 . If M = N , then the least value of 2 + 2 is equal to ______

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Solution

Q.
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 ______ . [2024]

Ans.
(5)

We have, $f (a) = \frac{4^{a}}{4^{a} + 2}$

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

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

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

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

$\Rightarrow M = \int_{f (a)}^{f (1 - a)} (1 - x) \sin^{4} [(1 - x) (1 - 1 + x)] d x$

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

$\Rightarrow \int_{f (a)}^{f (1 - a)} \sin^{4} [x (1 - x)] d x - \int_{f (a)}^{f (1 - a)} x \sin^{4} [x (1 - x)] d x$

$\Rightarrow M = N - M \Rightarrow 2 M = N \Rightarrow \frac{M}{N} = \frac{1}{2}$

$\Rightarrow \frac{M}{N} = \frac{β}{α} = \frac{1}{2} \Rightarrow β = 1, α = 2$

Then, the least value of $α^{2} + β^{2} = 4 + 1 = 5$

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