Question

The set of values of λ for which the function f(x) = $\frac{\lambda \sin x+6\cos x}{2\sin x+3\cos x}$ is strictly increasing, is ___________________.

Answer 1

The given function is $f\left(x\right)=\frac{\lambda \sin x+6\cos x}{2\sin x+3\cos x}$.

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

Differentiating both sides with respect to x, we get

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

$\Rightarrow f\text{'}\left(x\right)=\frac{2\lambda \sin x\cos x-12{\sin }^{2}x+3\lambda {\cos }^{2}x-18\sin x\cos x-2\lambda \sin x\cos x+3\lambda {\sin }^{2}x-12{\cos }^{2}x+18\sin x\cos x}{{\left(2\sin x+3\cos x\right)}^2}$

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

$\Rightarrow f\text{'}\left(x\right)=\frac{3\lambda -12}{{\left(2\sin x+3\cos x\right)}^2}$ [sin²x + cos²x = 1]

For f(x) to be strictly increasing,

$f\text{'}\left(x\right)>0$

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

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

$\Rightarrow 3\lambda >12$

$\Rightarrow \lambda >4$

Thus, the set of values of λ for which the function f(x) is strictly increasing is (4, ∞).


The set of values of λ for which the function f(x) = $\frac{\lambda \sin x+6\cos x}{2\sin x+3\cos x}$ is strictly increasing, is ____(4, ∞)____.