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let f(x)=-x^3+(b^3-b^2+b-1)/(b^2+3b+2) i...

let `f(x)=-x^3+(b^3-b^2+b-1)/(b^2+3b+2)` if `x` is `0` to `1` and `f(x)=2x-3` if `x` if `1` to `3`.All possible real values of `b` such that `f (x)` has the smallest value at `x=1` ,are

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The correct Answer is:
` b in (-2, -1 ) uu[1,, oo]`
Given, ` f(x) = {{:( -x^(3)+ (( b^(3) - b ^(2) + b - 1 ))/( (b ^(2) + 3 b + 2 ))",",, if 0 le x le 1 ) , ( 2x - 3",",, if 1 le x le 3 ):}`
is smallest at ` x = 1 `.
So, ` f(x)` is decreasing on `[0, 1]` and increasing on `[ 1, 3]`.
Here, `f (1) =- 1` is the smallest value at ` x = 1 `.
` therefore ` Its smallest value occur as
`underset ( x to 1 ^(-))(lim) f (x)= underset ( x to 1^(-)) ( lim) (-x ^(3)) + ((b^(3) - b ^(2) + b - 1 ))/( b^(2) + 3b + 2b )`
In order this value is not less than - 1, we must have
`" " (b^(3) - b ^(2) + b - 1 )/( b ^(2) + 3b + 2 ) ge 0`
`rArr " " (( b^(2) + 1) ( b - 1))/( ( b + 1 ) (b + 2)) ge 0`

` therefore " " b in (-2, -1 )uu [ 1, oo]`
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