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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    If m is the minimum value of k for which the function \[f(x)=x\sqrt{kx-{{x}^{2}}}\]is increasing in the interval\[\left[ 0,3 \right]\]and M is the maximum value of f in\[\left[ 0,3 \right]\]when \[k=m,\]then the ordered pair (m, M) is equal to: [JEE Main Held on 12-4-2019 Morning]

    A) \[(4,3\sqrt{2})\]

    B)               \[(4,3\sqrt{3})\]

    C) \[(3,3\sqrt{3})\]                      

    D) \[(5,3\sqrt{6})\]

    Correct Answer: B

    Solution :

    \[f(x)=x\sqrt{kx-{{x}^{2}}}\]           \[f'(x)=\frac{3kx-4{{x}^{2}}}{2\sqrt{kx-{{x}^{2}}}}\]
    For\[\uparrow f'(x)\ge 0\] \[kx-{{x}^{2}}\ge 0\] \[3kx-4{{x}^{2}}\ge 0\] \[4{{x}^{2}}-3kx\le 0\]
    \[{{x}^{2}}-kx\le 0\] \[4x(x-\frac{3k}{4})\le 0\]
    \[x(x-k)\le 0\]so\[x\in [0,3]\] \[3-\frac{3k}{4}\le 0\]
    \[+\text{v}e\] \[\]
    minimum value of k is \[\] \[f(x)=x\sqrt{kx-{{x}^{2}}}\] \[=3\sqrt{4\times 3-{{3}^{2}}}=3\sqrt{3},M=3\sqrt{3}\]


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