Let f:mathbbR→ mathbbR be a function defined by f(x)= log_√m √2( sinx- cosx)+m-2, for some m, such that the range of f is [0,2]. Then the value of m…

Given,

$f:\mathbb{R}\rightarrow \mathbb{R}$f:R→R be a function defined by $f(x)={\mathrm{log}}_{\sqrt{m}}{\sqrt{2}(\mathrm{sin}x-\mathrm{cos}x)+m-2}$f(x)=logm​​2​(sinx−cosx)+m−2, for some $m$m,

Also given the range of $f$f is $[0,2]$[0,2],

Now we know that,

$$-\sqrt{2}\leq \mathrm{sin}x-\mathrm{cos}x\leq \sqrt{2}$$−2​≤sinx−cosx≤2​

$$\Rightarrow -2\leq \sqrt{2}(\mathrm{sin}x-\mathrm{cos}x)\leq 2$$⇒−2≤2​(sinx−cosx)≤2

(Assuming $\sqrt{2}(\mathrm{sin}x-\mathrm{cos}x)=k$2​(sinx−cosx)=k)

$$\Rightarrow -2\leq k\leq 2\ldots (1)$$⇒−2≤k≤2…(1)

Now taking function $f(x)={\mathrm{log}}_{\sqrt{m}}(k+m-2)$f(x)=logm​​(k+m−2)

Given, $0\leq f(x)\leq 2$0≤f(x)≤2

$$\Rightarrow 0\leq {\mathrm{log}}_{\sqrt{m}}(k+m-2)\leq 2$$⇒0≤logm​​(k+m−2)≤2

$$\Rightarrow 1\leq k+m-2\leq m$$⇒1≤k+m−2≤m

$$\Rightarrow -m+3\leq k\leq 2\ldots (2)$$⇒−m+3≤k≤2…(2)

Now from equations (1)&(2), we get

$$-m+3=-2$$−m+3=−2

$$\Rightarrow m=5$$⇒m=5