Question

The greatest and least value of ${\log }_{\sqrt{2}}(\sin x-\cos x+3\sqrt{2})$ are respectively.

• A. $2$ and $1$
• B. $5$ and $3$
• C. $7$ and $5$
• D. $9$ and $7$

Answer 1

The correct option is B $5$ and $3$

Greatest Value of $sinx-cosx=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}$

least value $=-\sqrt{2}$

because maximum and minimum values of

$acosx+bsinx+c=c\pm \sqrt{a^2+b^2}$

so maximum value of $lo{g}_{\sqrt{2}}(sinx-cosx+3\sqrt{2})$

$=lo{g}_{\sqrt{2}}(\sqrt{2}+3\sqrt{2})=lo{g}_{\sqrt{2}}4\sqrt{2}$

$=lo{g}_{\sqrt{2}}(\sqrt{2}{)}^5$

$=5$

Minimum value of $lo{g}_{\sqrt{2}}(sinx-cosx+3\sqrt{2})$

$=lo{g}_{\sqrt{2}}\left(3\sqrt{2}\right)-\sqrt{2}=lo{g}_{\sqrt{2}}^{2\sqrt{2}}$

$=lo{g}_{\sqrt{2}}^{(\sqrt{2}{)}^3}$

$=3$