Question

If the function $f ( x )=\sin (\ln x )-2 \cos (\ln x)$ is increasing in the interval $\left( e ^\lambda, e ^\mu\right)$ then $\sqrt{5} \cos (\mu-\lambda)$ is equal to:

• A. $-\sqrt{5}$
• B. $\frac{-\sqrt{5}}{2}$
• C. 0
• D. $\sqrt{5}$

Answer 1

Answer: (A)

$f ^{\prime}( x )=\frac{\cos (\ln x )+2 \sin (\ln x )}{ x }=\frac{\sqrt{5}}{ x } \cdot\left(\cos (\ln x ) \frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}} \sin (\ln x )\right) $
$=\frac{\sqrt{5}}{ x } \cdot \sin (\ln x +\alpha) \text { where } \tan \alpha=\frac{1}{2}$
For increasing, $f^{\prime}(x)>0$
$\Rightarrow 0<\ln x +\alpha<\pi $
$\Rightarrow -\alpha<\ln x<\pi-\alpha $
$\Rightarrow n \pi-\alpha<\operatorname{lnx}< n \pi+\pi-\alpha$
$\Rightarrow e ^{ n \pi-\alpha}< x < e ^{ n \pi+\pi-\alpha} $
$\sqrt{5} \cos (\mu-\lambda)=\sqrt{5} \cos ( n \pi+\pi-\alpha-( n \pi-\alpha)=-\sqrt{5}$