The value of a such that f(x) = sqrt{3} sin x | vedclass

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(B) For the function $f(x)$ to be decreasing for all real values of $x$,its derivative $f'(x)$ must be less than or equal to $0$ for all $x \in \mathb

Vedclass · Accepted Answer

$a \geqslant 1$

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(B) For the function $f(x)$ to be decreasing for all real values of $x$,its derivative $f'(x)$ must be less than or equal to $0$ for all $x \in \mathbb{R}$.Given $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$.Differentiating with respect to $x$,we get:$f'(x) = \sqrt{3} \cos x + \sin x - 2a$.For $f(x)$ to be decreasing,$f'(x) \leqslant 0$ for all $x$.$\sqrt{3} \cos x + \sin x - 2a \leqslant 0$$\sqrt{3} \cos x + \sin x \leqslant 2a$.We know that the expression $A \cos x + B \sin x$ lies in the interval $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.Here,$A = \sqrt{3}$ and $B = 1$,so $\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.Thus,the maximum value of $\sqrt{3} \cos x + \sin x$ is $2$.For the inequality $2 \leqslant 2a$ to hold for all $x$,the maximum value of the left side must be less than or equal to $2a$.$2 \leqslant 2a \implies a \geqslant 1$.Therefore,the correct option is $B$.

The value of $a$ such that $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$ decreases for all real values of $x$ is given by:

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