Question

Let $f:R\to R$ be a function defined by $f(x)={\log }_{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is._____

• A. 3
• B. 5
• C. 4
• D. 2

Answer 1

Answer: (B)

Since,
$-\sqrt{2}\le \sin x-\cos x\le \sqrt{2}$
$\therefore -2\le \sqrt{2}(\sin x-\cos x)\le 2$
$(\text{ Assume }\sqrt{2}(\sin x-\cos x)=k)$
$-2\le k\le 2\dots \text{ (i) }$
$f(x)={\log }_{\sqrt{m}}(k+m-2)$
$\text{ Given, }$
$0\le f(x)\le 2$
$0\le {\log }_{\sqrt{m}}(k+m-2)\le 2$
$1\le k+m-2\le m$
$-m+3\le k\le 2\dots \text{ (ii) }$
From eq. (i) & (ii), we get $-m+3=-2$
$\Rightarrow m=5$