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Draw the graph of f(x) " maximum " {2 sin x, 1 - cos x}, x in (0, pi). Also find the range of g(x) " min " {2 sin x, 1 - cos x}, x in (0, pi) |
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Answer» Solution :Let us first DRAW graphs of `g_(1)(x) = 2 sin x` and `g_(2)(x) = 1 - cos x` `g_(1)(0) = g_(1)(pi) = 0, g_(1)(pi//2) = 2` `g_(2)(x) = sin x gt 0, AA x in (0, pi)` So `g_(2)' (x)` is increasing. `g_(2)(0) = 0, g_(2)(pi//2) = 1, g_(2)(pi) = 2` Graph of functions are as shown in the FOLLOWING figure.  Curves y = 2 sin x and y = 1 - cos x intersect when `4 sin^(2)x = (1 - cos x)^(2)` `rArr` `4(1 + cos x) = (1 - cos x)` `rArr` `4 + 4 cos x = 1 - cos x` `rArr` `cos x = -3//5` `rArr` `x = cos^(-1) (-3//5)` `THEREFORE` `f(x) = {{:(2sinx",", 0 LT x lt pi - "cos"^(-1)(3)/(5)),(1- cos x",", pi - "cos"^(-1)(3)/(5) lt x lt pi):}` Also the range of `g(x) = min {2 sin x, 1-cosx}` is `[0,1 - cos cos^(-1)(-(3)/(5))] -=[1, 8//5].` |
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