Function f(x) = frac{lambda sin x + 3 cos x}{ | vedclass

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(B) Given the function $f(x) = \frac{\lambda \sin x + 3 \cos x}{2 \sin x + 6 \cos x}$.To check for monotonicity,we find the derivative $f'(x)$ using t

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$\lambda > 1$

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(B) Given the function $f(x) = \frac{\lambda \sin x + 3 \cos x}{2 \sin x + 6 \cos x}$.To check for monotonicity,we find the derivative $f'(x)$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$.Here,$u = \lambda \sin x + 3 \cos x$ and $v = 2 \sin x + 6 \cos x$.$u' = \lambda \cos x - 3 \sin x$ and $v' = 2 \cos x - 6 \sin x$.$f'(x) = \frac{(\lambda \cos x - 3 \sin x)(2 \sin x + 6 \cos x) - (\lambda \sin x + 3 \cos x)(2 \cos x - 6 \sin x)}{(2 \sin x + 6 \cos x)^2}$.Expanding the numerator:$(\lambda \cos x - 3 \sin x)(2 \sin x + 6 \cos x) = 2\lambda \sin x \cos x + 6\lambda \cos^2 x - 6 \sin^2 x - 18 \sin x \cos x$.$(\lambda \sin x + 3 \cos x)(2 \cos x - 6 \sin x) = 2\lambda \sin x \cos x - 6\lambda \sin^2 x + 6 \cos^2 x - 18 \sin x \cos x$.Subtracting these:Numerator $= (6\lambda \cos^2 x - 6 \sin^2 x) - (-6\lambda \sin^2 x + 6 \cos^2 x) = 6\lambda(\cos^2 x + \sin^2 x) - 6(\sin^2 x + \cos^2 x) = 6\lambda - 6 = 6(\lambda - 1)$.Thus,$f'(x) = \frac{6(\lambda - 1)}{(2 \sin x + 6 \cos x)^2}$.For the function to be monotonically increasing,$f'(x) > 0$ for all $x$ where the function is defined.Since the denominator is a square,it is always positive. Therefore,we require $6(\lambda - 1) > 0$,which implies $\lambda - 1 > 0$,or $\lambda > 1$.

Function $f(x) = \frac{\lambda \sin x + 3 \cos x}{2 \sin x + 6 \cos x}$ is monotonic increasing,then :

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