The function f(x) = frac{lambda sin x + 6 cos | vedclass

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

Question

(C) Given the function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.For the function to be increasing,we must have $f'(x) \geq 0$.Us

Vedclass · Accepted Answer

$\lambda \geq 4$

Vedclass · Answer

(C) Given the function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$.For the function to be increasing,we must have $f'(x) \geq 0$.Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$,we calculate $f'(x)$:$f'(x) = \frac{(\lambda \cos x - 6 \sin x)(2 \sin x + 3 \cos x) - (\lambda \sin x + 6 \cos x)(2 \cos x - 3 \sin x)}{(2 \sin x + 3 \cos x)^2}$.Expanding the numerator:Numerator $= (2\lambda \sin x \cos x + 3\lambda \cos^2 x - 12 \sin^2 x - 18 \sin x \cos x) - (2\lambda \sin x \cos x - 3\lambda \sin^2 x + 12 \cos^2 x - 18 \sin x \cos x)$.Simplifying the numerator:Numerator $= 3\lambda \cos^2 x - 12 \sin^2 x + 3\lambda \sin^2 x - 12 \cos^2 x$.Numerator $= 3\lambda(\sin^2 x + \cos^2 x) - 12(\sin^2 x + \cos^2 x) = 3\lambda - 12$.Since the denominator $(2 \sin x + 3 \cos x)^2$ is always positive,$f'(x) \geq 0$ implies $3\lambda - 12 \geq 0$.Therefore,$3\lambda \geq 12$,which gives $\lambda \geq 4$.

The function $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is increasing,if

Similar Questions

The least value of $k$ for which the function $f(x) = {x^2} + kx + 1$ is an increasing function in the interval $1 \leq x \leq 2$ is

The length of the longest interval in which the function $f(x) = 3 \sin x - 4 \sin^3 x$ is increasing is:

The function $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$ is

The function $f(x) = (x - 2)|x - 3|$ is monotonically increasing in:

Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is $(i)$ increasing $(ii)$ decreasing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module