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(B) First,observe that $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2} = \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x+2} = \frac{4^x}{4^x+2}

Vedclass · Accepted Answer

$5$

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(B) First,observe that $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2} = \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x+2} = \frac{4^x}{4^x+2} + \frac{4}{4+2 \cdot 4^x} = \frac{4^x}{4^x+2} + \frac{2}{2+4^x} = \frac{4^x+2}{4^x+2} = 1.$Let $I = \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) dx.$ Using the property $\int_A^B g(x) dx = \int_A^B g(A+B-x) dx,$ and noting that $A+B = f(a) + f(1-a) = 1,$ we have:$M = \int_{f(a)}^{f(1-a)} (1-x) \sin^4((1-x)(1-(1-x))) dx = \int_{f(a)}^{f(1-a)} (1-x) \sin^4((1-x)x) dx.$Thus,$M = \int_{f(a)}^{f(1-a)} \sin^4(x(1-x)) dx - \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) dx.$This implies $M = N - M,$ which simplifies to $2M = N.$Given $\alpha M = \beta N,$ we have $\alpha M = \beta (2M),$ so $\alpha = 2\beta.$Since $\alpha, \beta \in N,$ the smallest values are $\beta = 1$ and $\alpha = 2.$The value of $\alpha^2 + \beta^2 = 2^2 + 1^2 = 4 + 1 = 5.$

Let $f: R \rightarrow R$ be a function defined by $f(x)=\frac{4^x}{4^x+2}$ and $M=\int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) dx,$ $N=\int_{f(a)}^{f(1-a)} \sin^4(x(1-x)) dx;$ $a \neq \frac{1}{2}.$ If $\alpha M=\beta N,$ $\alpha, \beta \in N,$ then the least value of $\alpha^2+\beta^2$ is equal to $.....$

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