Question

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