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Let f: \mathbb{R} \rightarrow \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

Option: 1

5


Option: 2

4


Option: 3

3


Option: 4

2


Answers (1)

\begin{aligned} & \because-\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2} \\ & \Rightarrow-2 \leq \sqrt{2}(\sin x-\cos x) \leq 2 \\ & \Rightarrow m-4 \leq \sqrt{2}(\sin x-\cos x)+m-2 \leq m \\ & \Rightarrow \log _{\sqrt{\mathrm{m}}}^{(\mathrm{m}-4)} \leq \log _{\sqrt{\mathrm{m}}}^{\{\sqrt{2}(\sin x-\cos x)+\mathrm{m}-2\}} \leq \log _{\sqrt{\mathrm{m}}}^{\mathrm{m}} \\ & \quad \Downarrow \\ & \quad 0 \\ & \Rightarrow \log _{\sqrt{\mathrm{m}}}^{(\mathrm{m}-4)}=0 \\ & \Rightarrow \mathrm{m}=5 \end{aligned}

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Sumit Saini

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