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

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

$$\mathrm{Let}\:{k}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\sqrt{{x}\:−\:\mathrm{3}}\:+\:\sqrt{\mathrm{6}\:−\:{x}}\:\geqslant\:{k}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{solution}\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{k} \\ $$$$\mathrm{is} \\ $$

Commented by mrW1 last updated on 20/Sep/17

k_(max) =(√6)

$$\mathrm{k}_{\mathrm{max}} =\sqrt{\mathrm{6}} \\ $$

Commented by Tinkutara last updated on 21/Sep/17

How?

$$\mathrm{How}? \\ $$

Commented by Tinkutara last updated on 21/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by mrW1 last updated on 21/Sep/17

f(x)=(√(x−3))+(√(6−x))  3≤x≤6  f(3)=(√3)  f(6)=(√3)  f′(x)=(1/(2(√(x−3))))−(1/(2(√(6−x))))  f′(x)=0⇒(√(6−x))=(√(x−3))⇒x=(9/2)  f((9/2))=(√((9/2)−3))+(√(6−(9/2)))=(√6)>(√3)  f′′(x)=−(1/(4(x−3)(√(x−3))))−(1/(4(6−x)(√(6−x))))  f′′((9/2))=−(1/(4×(3/2)×(√(3/2))))−(1/(4×(3/2)×(√(3/2))))=−(2/(3(√6)))<0  ⇒f(x) has an absolute max. at x=(9/2)  which is (√6).  ⇒(√3)≤ f(x)=(√(x−3))+(√(6−x))≤(√6)    or f(x)≥k has a solution only if k≤(√6).

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}−\mathrm{3}}+\sqrt{\mathrm{6}−\mathrm{x}} \\ $$$$\mathrm{3}\leqslant\mathrm{x}\leqslant\mathrm{6} \\ $$$$\mathrm{f}\left(\mathrm{3}\right)=\sqrt{\mathrm{3}} \\ $$$$\mathrm{f}\left(\mathrm{6}\right)=\sqrt{\mathrm{3}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}−\mathrm{x}}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\sqrt{\mathrm{6}−\mathrm{x}}=\sqrt{\mathrm{x}−\mathrm{3}}\Rightarrow\mathrm{x}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{f}\left(\frac{\mathrm{9}}{\mathrm{2}}\right)=\sqrt{\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{3}}+\sqrt{\mathrm{6}−\frac{\mathrm{9}}{\mathrm{2}}}=\sqrt{\mathrm{6}}>\sqrt{\mathrm{3}} \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{x}−\mathrm{3}\right)\sqrt{\mathrm{x}−\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{6}−\mathrm{x}\right)\sqrt{\mathrm{6}−\mathrm{x}}} \\ $$$$\mathrm{f}''\left(\frac{\mathrm{9}}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{4}×\frac{\mathrm{3}}{\mathrm{2}}×\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}}−\frac{\mathrm{1}}{\mathrm{4}×\frac{\mathrm{3}}{\mathrm{2}}×\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}}=−\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{6}}}<\mathrm{0} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{an}\:\mathrm{absolute}\:\mathrm{max}.\:\mathrm{at}\:\mathrm{x}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\sqrt{\mathrm{6}}. \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\leqslant\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}−\mathrm{3}}+\sqrt{\mathrm{6}−\mathrm{x}}\leqslant\sqrt{\mathrm{6}} \\ $$$$ \\ $$$$\mathrm{or}\:\mathrm{f}\left(\mathrm{x}\right)\geqslant\mathrm{k}\:\mathrm{has}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{only}\:\mathrm{if}\:\mathrm{k}\leqslant\sqrt{\mathrm{6}}. \\ $$

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