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The value of $K$ in order that $f(x)=\sin x-\cos x-K x+5$ decreases for all positive real values of $x$ is given by
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Verified Answer
The correct answer is:
$K>\sqrt{2}$
We have,
$\begin{aligned} f(x) &=\sin x-\cos x-K x+5 \\ \Rightarrow f^{\prime}(x) &=\cos x+\sin x-K \end{aligned}$
For decreasing, $f^{\prime}(x) < 0$ $\Rightarrow \cos x+\sin x-K < 0$
$\begin{array}{ll}\Rightarrow & K>\cos x+\sin x \\ \Rightarrow & \quad K>\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right] \\ \Rightarrow & K>\sqrt{2} \sin \left(\frac{\pi}{4}+x\right) \\ \therefore \quad & K>\sqrt{2}\end{array}$
$\begin{aligned} f(x) &=\sin x-\cos x-K x+5 \\ \Rightarrow f^{\prime}(x) &=\cos x+\sin x-K \end{aligned}$
For decreasing, $f^{\prime}(x) < 0$ $\Rightarrow \cos x+\sin x-K < 0$
$\begin{array}{ll}\Rightarrow & K>\cos x+\sin x \\ \Rightarrow & \quad K>\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right] \\ \Rightarrow & K>\sqrt{2} \sin \left(\frac{\pi}{4}+x\right) \\ \therefore \quad & K>\sqrt{2}\end{array}$
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