JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 12)
Let $$f:\mathbb{R}\to\mathbb{R}$$ 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 _________
4
3
5
2
Explanation
We know that $\sin x-\cos x \in[-\sqrt{2}, \sqrt{2}]$
$$ \begin{aligned} & \log _{\sqrt{M}}(\sqrt{2}(\sin x-\cos ) +M-2) \\\\ &\quad\quad\in {\left[\log _{\sqrt{M}}(M-4), \log _{\sqrt{M}} M\right] } \end{aligned} $$
$\Rightarrow \log _{\sqrt{M}}(M-4)=0 \Rightarrow M=5$
$$ \begin{aligned} & \log _{\sqrt{M}}(\sqrt{2}(\sin x-\cos ) +M-2) \\\\ &\quad\quad\in {\left[\log _{\sqrt{M}}(M-4), \log _{\sqrt{M}} M\right] } \end{aligned} $$
$\Rightarrow \log _{\sqrt{M}}(M-4)=0 \Rightarrow M=5$
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