Let f: ℝ → ℝ be a function defined by f(x) = _ m \ 2 ( x - x) + m -...

Given,

$f: \mathbb{R} \to \mathbb{R}$ be a function defined by $f\left(x\right) = \log_{\sqrt{m}}\left\{\sqrt{2}\left(\sin x - \cos x\right) + m - 2\right\}$, for some $m$,

Also given the range of $f$ is \left[0, 2\right]$,</p><p>Now we know that,</p><p>$- \sqrt{2} \le \sin x - \cos x \le \sqrt{2}$</p><p>$⇒ - 2 \le \sqrt{2}\left(\sin x - \cos x\right) \le 2$</p><p>(Assuming$\sqrt{2}\left(\sin x - \cos x\right) = k$)</p><p>$⇒ - 2 \le k \le 2 \ldots \left(1\right)$</p><p>Now taking function$f\left(x\right) = \log_{\sqrt{m}}\left(k + m - 2\right)$</p><p>Given,$0 \le f\left(x\right) \le 2$</p><p>$⇒ 0 \le \log_{\sqrt{m}}\left(k + m - 2\right) \le 2$</p><p>$⇒ 1 \le k + m - 2 \le m$</p><p>$⇒ - m + 3 \le k \le 2 \ldots \left(2\right)$</p><p>Now from equations \left(1\right) \text{ and } \left(2\right)$, we get

$- m + 3 = - 2$

$⇒ m = 5$