The value of K such that f(x) = sin x - cos x | vedclass

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Question

(B) Given $f(x) = \sin x - \cos x - Kx + 5$.Differentiating with respect to $x$,we get $f'(x) = \cos x + \sin x - K$.For the function to be decreasing

Vedclass · Accepted Answer

$K \geq 1$

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(B) Given $f(x) = \sin x - \cos x - Kx + 5$.Differentiating with respect to $x$,we get $f'(x) = \cos x + \sin x - K$.For the function to be decreasing for all $x > 0$,we must have $f'(x) \leq 0$ for all $x$.This implies $\cos x + \sin x - K \leq 0$,or $K \geq \cos x + \sin x$.We know that the maximum value of $\cos x + \sin x$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$.Therefore,for $K \geq \cos x + \sin x$ to hold for all $x$,$K$ must be greater than or equal to the maximum value of the expression $\cos x + \sin x$.Thus,$K \geq \sqrt{2}$.

The value of $K$ such that $f(x) = \sin x - \cos x - Kx + 5$ decreases for all positive real values of $x$ is given by

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