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Find the minimum value of f(x) = (x^2sin^2x + 4)/xsinx, where x ∈ (0, π/2)

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in Differential equations by (50.3k points)

Find the minimum value of f(x)  = (x2sin2x + 4)/xsinx, where x ∈ (0, π/2)

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We have, f(x) = (x2sin2x + 4)/xsinx

= xsinx + 4/xsinx ≤ 4

Hence, the minimum values of f(x) is 4

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