JEE Main 2023 (Online) 25th January Evening ShiftQuestion 1
Let $f:\mathbb R\to\mathbb R$ be a function defined by $f(x)=\log_{\sqrt m}\left\{\sqrt2(\sin x-\cos x)+m-2\right\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is :
Functions
Solution
$\sqrt2(\sin x-\cos x)\in[-2,2]$, so the argument ranges over $[m-4,m]$. Range $[0,2]$ needs $m-4=1$ and $\log_{\sqrt m}m=2$; $m=5$.