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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 decreasin

Vedclass · Accepted Answer

$K \geq 1$

Vedclass · Answer

(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

Similar Questions

The function $f(x) = \tan x - x$ is:

Find the intervals in which the function $f(x) = (x+1)^{3}(x-3)^{3}$ is strictly increasing or strictly decreasing.

If $F(x) = \int_{0}^{x} \frac{\cos t}{1+t^{2}} dt$, where $0 \leq x \leq 2\pi$, then which of the following is true?

The function $\frac{a\sin x + b\cos x}{c\sin x + d\cos x}$ is decreasing,if

Find the intervals in which the function given by $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$ is $(a)$ increasing $(b)$ decreasing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module