Let-k-be-a-real-number-such-that-the-inequality-x-3-6-x-k-has-a-solution-then-the-maximum-value-of-k-is-

Question Number 21313 by Tinkutara last updated on 20/Sep/17

Let k be a real number such that the

inequality \sqrt{x - 3} + \sqrt{6 - x} ⩾ k has a

solution then the maximum value of k

is

Commented by mrW1 last updated on 20/Sep/17

k_{\max} = \sqrt{6}

Commented by Tinkutara last updated on 21/Sep/17

How ?

Commented by Tinkutara last updated on 21/Sep/17

Thank you very much Sir!

Commented by mrW1 last updated on 21/Sep/17

f (x) = \sqrt{x - 3} + \sqrt{6 - x}

3 ⩽ x ⩽ 6

f (3) = \sqrt{3}

f (6) = \sqrt{3}

f^{'} (x) = \frac{1}{2 \sqrt{x - 3}} - \frac{1}{2 \sqrt{6 - x}}

f^{'} (x) = 0 \Rightarrow \sqrt{6 - x} = \sqrt{x - 3} \Rightarrow x = \frac{9}{2}

f (\frac{9}{2}) = \sqrt{\frac{9}{2} - 3} + \sqrt{6 - \frac{9}{2}} = \sqrt{6} > \sqrt{3}

f^{″} (x) = - \frac{1}{4 (x - 3) \sqrt{x - 3}} - \frac{1}{4 (6 - x) \sqrt{6 - x}}

f^{″} (\frac{9}{2}) = - \frac{1}{4 \times \frac{3}{2} \times \sqrt{\frac{3}{2}}} - \frac{1}{4 \times \frac{3}{2} \times \sqrt{\frac{3}{2}}} = - \frac{2}{3 \sqrt{6}} < 0

\Rightarrow f (x) has an absolute \max . at x = \frac{9}{2}

which is \sqrt{6} .

\Rightarrow \sqrt{3} ⩽ f (x) = \sqrt{x - 3} + \sqrt{6 - x} ⩽ \sqrt{6}

or f (x) ⩾ k has a solution only if k ⩽ \sqrt{6} .

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