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The value of \(a\) in order that \(f(x)=\sin x-\cos x-a x+b\) decreases for all real values is given by
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The correct answer is:
\(a \geq \sqrt{2}\)
We have; \(f(x)=\sin x-\cos x-a x+b\)
\(\Rightarrow f^{\prime}(x)=\cos x+\sin x-a\)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) < 0 ~\forall~ \mathrm{x} \in \mathrm{R}\)
\(\Rightarrow(\cos x+\sin x) < a ~\forall~ x \in R\)
As the max. value of \((\cos x+\sin x)\) is \(\sqrt{2}\)
The above is possible when \(a \geq \sqrt{2}\)
\(\Rightarrow f^{\prime}(x)=\cos x+\sin x-a\)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) < 0 ~\forall~ \mathrm{x} \in \mathrm{R}\)
\(\Rightarrow(\cos x+\sin x) < a ~\forall~ x \in R\)
As the max. value of \((\cos x+\sin x)\) is \(\sqrt{2}\)
The above is possible when \(a \geq \sqrt{2}\)
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