Question

Function $f\left(x\right)=\frac{[\lambda \sin x+6\cos x]}{[2\sin x+3\cos x]}$ is monotonic increasing, if

• A. $\lambda >1$
• B. $\lambda <1$
• C. $\lambda <4$
• D. $\lambda >4$

Answer 1

The correct option is D

$\lambda >4$

Explanation for the correct option:

Step 1. Find the value of $\lambda$:

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

On differentiating it with respect to $x$, we get

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

$\mathbf{\left[}\mathbf{\because }\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}}\mathbf{\right]}$

$\Rightarrow$ $f\text{'}\left(x\right)=\frac{3\lambda ({\sin }^{2}x+{\cos }^{2}x)–12({\sin }^{2}x+{\cos }^{2}x)}{{\left(2\sin x+3\cos x\right)}^2}$

Step 2. The function is monotonic increasing, if $f\prime \left(x\right)>0$

$\Rightarrow$$3\lambda ({\sin }^{2}x+{\cos }^{2}x)-12({\sin }^{2}x+{\cos }^{2}x)>0$

$\Rightarrow$ $3\lambda -12>0$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\mathbf{\therefore }\mathit{\lambda }\mathbf{>}\mathbf{4}$

Hence, Option ‘D’ is Correct.