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Let f(x) = 4x^(2)-4ax+a^(2)-2a+2 and the global minimum value of f(x) for x in [0,2] is equal to 3 The values of a for which f(x) is monotonic for x in [0,2] |
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Answer» `a LE` 0 or a `ge 4` vertex of this parabola is `((a)/(2),2-2a)` caseI:`0 lt a/2 lt 2` In this case f(x) will attain the minium VALUE at `x=a/2` .Thus `f(a/2)^(3)`  or `3=-2a+2 or a =-1/2` (rejected) Case II: `(a)/(2)ge2` In this f(x) attains global minimum value at x =2 thus f(2)=3 `3=16-8a+a^(2)-2a+2 or a =5 pm sqrt(10)` thus `a=5+sqrt(10)` Case III: `(a)/(2)ge0` In this case f(x) attains the global minumum value at x =0 thus f(0)=3 `therefore3=a^(2)-2a+2 or a =1 pm` Thus a =1-`sqrt(2)` Hence the permissibel VALUES of a are `1-sqrt(2) and 2+sqrt(10)` f(x) =`4x^(2)-49x+a^(2)-2a+2`is monotonic in [0,2] Hence the point of minima of funciton should not lie in [0,2] Now f(x) =0 or 8x-4a=0 or x `=a//2` `(a)/(2)in [0,2] or a in [0,4]` For f(x) to be monotomic [0,2], and[0,4] i/e`ale0` or age4` |
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