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If m is the minimu value of k for which the function `f(x)=xsqrt(kx-x^(2))` is increasing in the interval [0,3] and M is the maximum value of f in the inverval [0,3] when h=m, then the ordered pair (m,M) is equal to

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The correct Answer is:
4
`f'(n)=(x)/(2sqrt(kn-x^(2))).(k-2n)+sqrt(kx+x^(2))=0`
`rArr" "kx-2x^(2)+2(kx-x^(2))=0" "rArr" "3kx-4x^(2)=0" "rArr" "x=0" or "x=(3k)/(4)=3`
`k=4" "rArr" Max value of "f(x)=3sqrt3`
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