Let f : R rightarrow R be a function defined | vedclass

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(A) We know that $-\sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$.Multiplying by $\sqrt{2}$,we get $-2 \leq \sqrt{2}(\sin x - \cos x) \leq 2$.Let $k = \

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$5$

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(A) We know that $-\sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$.Multiplying by $\sqrt{2}$,we get $-2 \leq \sqrt{2}(\sin x - \cos x) \leq 2$.Let $k = \sqrt{2}(\sin x - \cos x)$,so $-2 \leq k \leq 2$.The function is $f(x) = \log_{\sqrt{m}}(k + m - 2)$.Given the range of $f$ is $[0, 2]$,we have $0 \leq \log_{\sqrt{m}}(k + m - 2) \leq 2$.This implies $(\sqrt{m})^0 \leq k + m - 2 \leq (\sqrt{m})^2$,which simplifies to $1 \leq k + m - 2 \leq m$.Solving for $k$,we get $3 - m \leq k \leq 2$.Comparing this with $-2 \leq k \leq 2$,we equate the lower bounds: $3 - m = -2$.Thus,$m = 5$.

Let $f : R \rightarrow R$ 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 $............$

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