Function, f (x) = λsinx+6cosx / 2sinx+3cosx is monotonic increasing, if

The correct option is λ > 4.

To find the condition under which the function is monotonically increasing, we need to analyze its first derivative. A function $f\left(x\right)$ is monotonically increasing if its derivative $f^{\prime }\left(x\right)>0$ for all points in its domain.

The given function is:

$$f\left(x\right)=\frac{\lambda \sin x+6\cos x}{2\sin x+3\cos x}$$

We can differentiate $f\left(x\right)$ with respect to $x$ using the quotient rule, which states that for a fraction $\frac{u}{v}$, the derivative is $\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$.

Let's define the numerator and denominator:

$u=\lambda \sin x+6\cos x\Rightarrow u^{\prime }=\lambda \cos x-6\sin x$

$v=2\sin x+3\cos x\Rightarrow v^{\prime }=2\cos x-3\sin x$

Now, applying the quotient rule:

$$f^{\prime }\left(x\right)=\frac{(\lambda \cos x-6\sin x)(2\sin x+3\cos x)-(\lambda \sin x+6\cos x)(2\cos x-3\sin x)}{{(2\sin x+3\cos x)}^2}$$

Expanding the terms in the numerator:

$$\text{Numerator}=(2\lambda \sin x\cos x+3\lambda {\cos }^{2}x-12{\sin }^{2}x-18\sin x\cos x)-(2\lambda \sin x\cos x-3\lambda {\sin }^{2}x+12{\cos }^{2}x-18\sin x\cos x)$$

Simplifying the numerator by combining like terms:

$$\text{Numerator}=3\lambda {\cos }^{2}x-12{\sin }^{2}x+3\lambda {\sin }^{2}x-12{\cos }^{2}x$$

$$\text{Numerator}=3\lambda ({\sin }^{2}x+{\cos }^{2}x)-12({\sin }^{2}x+{\cos }^{2}x)$$

Using the trigonometric identity ${\sin }^{2}x+{\cos }^{2}x=1$, the numerator simplifies directly to:

$$\text{Numerator}=3\lambda -12$$

Thus, the derivative of the function is:

$$f^{\prime }\left(x\right)=\frac{3\lambda -12}{{(2\sin x+3\cos x)}^2}$$

For $f\left(x\right)$ to be monotonic increasing, we require $f^{\prime }\left(x\right)>0$. Since the denominator ${(2\sin x+3\cos x)}^2$ is always positive for all values where the function is defined, the sign of $f^{\prime }\left(x\right)$ is entirely determined by the numerator.

Therefore, we set the numerator to be greater than zero:

$$3\lambda -12>0$$

$$3\lambda >12$$

$$\lambda >4$$

Consequently, the function is monotonic increasing if $\lambda >4$.