1.

Let f : R to R be a differentiable function given by f(x) =x^(3)-3x + 2020. If g(x) is a continuous function defined by g(x) ={{:("Minimum" f(t),0 le t le x, 0 le x le 1),("Maximum" f(t), 1 lt t le x, 1 lt x le 2):} and m and M be the least and the greatest value of g(x) on [0,2] then which one of the following is correct?

Answer»

M-m=2
m=2020
M=2022
m=2019

Solution :`f(x) = 3X^(2) -3`
For `0 le x le 1`, f(x) is STRICTLY DECREASING
`rArr g(x) =x^(2) - 3x + 2020, 0 le x le 1`
For `1 lt x le 2`, f(x) is strictly increasing `rArr g(x) = xp^(3) - 3x + 2020, 1 lt x le 2`


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