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The least value of the function (x) = 2cosx + x in the closed interval [0,π/2] is :(a) 2(b) π/6 + √3(c) π/2(d) The least value does not exist. |
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Answer» Option : (c) f(x) = 2cosx + x, x ∈ [0,π/2] f'(x) = −2sinx + 1 Letf′(x) = 0 ⇒ x = \(\frac{\pi}{6}\) ∈ [0,\(\frac{\pi}{2}\)] f(0) = 2 f (\(\frac{\pi}{6}\)) = \(\frac{\pi}{6}\) + √3 f (\(\frac{\pi}{2}\)) = \(\frac{\pi}{2}\) ⇒ least value of f(x) is \(\frac{\pi}{2}\) at x = \(\frac{\pi}{2}\) |
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