If f(x)=sqrt{3} sin x-cos x-2 a x+b decreases | vedclass

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

Question

(A) Given the function $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$.To find the condition for $f(x)$ to be decreasing,we first find its derivative $f^{\prime

Vedclass · Accepted Answer

$a \geq 1$

Vedclass · Answer

(A) Given the function $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$.To find the condition for $f(x)$ to be decreasing,we first find its derivative $f^{\prime}(x)$.$f^{\prime}(x)=\sqrt{3} \cos x+\sin x-2 a$.We can rewrite this as $f^{\prime}(x)=2 \left( \frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x \right)-2 a$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f^{\prime}(x)=2 \sin \left(x+\frac{\pi}{3}\right)-2 a$.For $f(x)$ to decrease for all $x$,we must have $f^{\prime}(x) \leq 0$ for all $x$.$2 \sin \left(x+\frac{\pi}{3}\right)-2 a \leq 0$.$2 a \geq 2 \sin \left(x+\frac{\pi}{3}\right)$.$a \geq \sin \left(x+\frac{\pi}{3}\right)$.Since the maximum value of $\sin \left(x+\frac{\pi}{3}\right)$ is $1$,for the inequality to hold for all $x$,we must have $a \geq 1$.

If $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b$ decreases for all values of $x$,then

Similar Questions

In which of the following intervals does $f(x) = 2x^3$ increase less rapidly than $g(x) = 9x^2 - 12x + 6$?

If $f(x) = \frac{k \sin x + 2 \cos x}{\sin x + \cos x}$ is strictly increasing for all real values of $x$,then

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

If $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1, x \neq 0$ and $y = 5x^2 f(x)$,then $y$ is strictly increasing in

Let $f(x) = xe^{x(1 - x)}$. Then $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module