Question: If the function $f(x)=\sin(\ln x)-2\cos(\ln x)$ is increasing in the interval $(e^\lambda, e^\mu)$ then $\sqrt 5\cos(\mu-\lambda)$ is equal to __?
My approach
- I took the first derivative to check the points of changing slope ( by putting $y=\ln x$)
- the points came out as $y=\tan^{-1}(-\frac{1}{2})$ and graph in desmos shows something else so am I supposed to take a double derivative and solve the question or is this approach right https://www.desmos.com/calculator/aimvhqe78f
The question is ill-posed, because if a function is increasing on $(a,b)$, then it is also increasing on $(a',b')$, for all $a',b'$ such that $a\le a'<b'\le b$.
If we amend the question by stating that $e^\lambda$ is a local minimum and $e^\mu$ is a local maximum, then it becomes answerable.
The function $f$ is increasing on $(e^\lambda,e^\mu)$ if and only if the function $g(x)=\sin x-2\cos x$ is increasing on $(\lambda,\mu)$. This holds because the natural logarithm is an increasing function.
Since $g'(x)=\cos x+2\sin x$ vanishes at $x=\arctan(-1/2)+2k\pi$ and at $x=\pi+\arctan(-1/2)+2k\pi$, what is $\mu-\lambda$?