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, $f^{\prime }(x)>0$
$\Rightarrow 0<\ln x+\alpha <\pi$
$\Rightarrow -\alpha <\ln x<\pi -\alpha$
$\Rightarrow n\pi -\alpha <\mathrm{lnx}<n\pi +\pi -\alpha$
$\Rightarrow e^{n\pi -\alpha }<x<e^{n\pi +\pi -\alpha }$
$\sqrt{5}\cos (\mu -\lambda )=\sqrt{5}\cos (n\pi +\pi -\alpha -(n\pi -\alpha )=-\sqrt{5}$