Let k and K be the minimum and the maximum values of the function `f(x) = ((1+x)^(0.6))/(1+x^(0.6)`, and `x in [0,1]` respectively,then the ordered pair (k, K) is equal to

Options

A. `(2^(-0.4),1)`
B. `(2^(-0.4),2^(0.6))`
C. `(2^(-0.6),1)`
D. `(1,2^(0.6))`

Answer

A

Explanation

Given `f(x)=((1+x)^(3//5))/(1+x^(3//5)),x epsilon[0,1]`
`f(x)=((1+x^(3//5))3/5(1+x)^(-2//5)-3/5(1+x)^(3//5)x^(-2//5))/((1+x^(3//5))^(2))`
`impliesf'(x)=(3/5[(1+x^(3//5))/((1+x)^(2//5))-((1+x)^(3//5))/(x^(2//5))])/((1+x^(3//5))^(2))`
`implies f'(x)=3/5[(x^(2//5)+x-1-x)/((1+x)^(2//5)x^(2//5)(1+x^(3//5))^(2))]`
`implies f'(x)=3/5[(x^(2//5)-1)/((1+x)^(2//5)x^(2//5)(1+x^(3//5))^(2))]`
`:'x^(2//5)-1lt0` when `x epsilon[0,1]impliesf'(x)lt0`
Hence given function is strictly decreasing in`x epsilon[0,1]`
Maximum value `K=f(0)=1`
Minimum value `k=f(1)=(2^(3//5))/2=2^(-0.4)`
Thus, the ordered pairis `(k,K)=(2^(-0.4),1)`.