Question

If the function $f\left(x\right)=\sin (\log x)-\cos (\log x)$ strictly increase in the interval $(e^{\lambda },e^{\mu })$ then the value of $-500\cos (\mu -\lambda )$.

Answer 1

Hint: We will first find the differentiation of the given function as we have given that the function strictly increases in the interval. Form that condition we will find the range of $x$ and we equate the result range with the given range to find the values of $\lambda ,\mu$ from that values we can find the required value of $-500\cos (\mu -\lambda )$

Complete step by step answer:
Given that, $f\left(x\right)=\sin (\log x)-\cos (\log x)$ strictly increase in the interval $(e^{\lambda },e^{\mu })$, then
$f^{{}^{\prime }}\left(x\right)>0$
The value of $f^{{}^{\prime }}\left(x\right)$ is
$\begin{array}{rl}& f^{{}^{\prime }}\left(x\right)={\displaystyle \frac{d}{dx}}[\sin (\log x)-\cos (\log x)]\\ & ={\displaystyle \frac{d}{dx}}[\sin (\log x)]-{\displaystyle \frac{d}{dx}}[\cos (\log x)]\end{array}$
Use the formulas ${\displaystyle \frac{d}{dx}}(\sin x)=\cos x$, ${\displaystyle \frac{d}{dx}}(\cos x)=-\sin x$ in the above equation, then
$f^{{}^{\prime }}\left(x\right)=\cos (\log x).{\displaystyle \frac{d}{dx}}(\log x)+\sin (\log x){\displaystyle \frac{d}{dx}}(\log x)$
Use the formula ${\displaystyle \frac{d}{dx}}(\log x)={\displaystyle \frac{1}{x}}$, then
$\begin{array}{rl}& f^{{}^{\prime }}\left(x\right)=\cos (\log x).{\displaystyle \frac{1}{x}}+\sin (\log x).{\displaystyle \frac{1}{x}}\\ & ={\displaystyle \frac{1}{x}}[\cos (\log x)+\sin (\log x)]\end{array}$
if the function $f\left(x\right)$ strictly increase in the interval $(e^{\lambda },e^{\mu })$, then
$\begin{array}{rl}& f^{{}^{\prime }}\left(x\right)>0\\ & {\displaystyle \frac{1}{x}}[\sin (\log x)+\cos (\log x)]>0\\ & \sin (\log x)+\cos (\log x)>0\end{array}$
Multiply and divide by $\sqrt{2}$ in the above expression, then we have
$\sqrt{2}({\displaystyle \frac{1}{\sqrt{2}}}\sin (\log x)+{\displaystyle \frac{1}{\sqrt{2}}}\cos (\log x))>0$
Substituting $\sin {\displaystyle \frac{\pi }{4}}=\cos {\displaystyle \frac{\pi }{4}}={\displaystyle \frac{1}{\sqrt{2}}}$in the above expression, then
$\cos {\displaystyle \frac{\pi }{4}}.\sin (\log x)+\sin {\displaystyle \frac{\pi }{4}}\cos (\log x)>0$
Using the formula $\sin x.\cos y+\cos x.\sin y=\sin (x+y)$ in the above expression, then we have
$\sin ({\displaystyle \frac{\pi }{4}}+\log x)>0$
The graph of $y=\sin x$ is given below

Form the above equation we have value of $\sin x>0$ for $0<x<\pi$, so the value of $\sin ({\displaystyle \frac{\pi }{4}}+\log x)>0$
For
$\begin{array}{rl}& 0<{\displaystyle \frac{\pi }{4}}+\log x<\pi \\ & -{\displaystyle \frac{\pi }{4}}<\log x<\pi -{\displaystyle \frac{\pi }{4}}\\ & -{\displaystyle \frac{\pi }{4}}<\log x<{\displaystyle \frac{3\pi }{4}}\\ & e^{-{\displaystyle \frac{\pi }{4}}}<x<e^{{\displaystyle \frac{3\pi }{4}}}\end{array}$
$\therefore$$x\in (e^{-{\displaystyle \frac{\pi }{4}}},e^{{\displaystyle \frac{3\pi }{4}}})$
But given that $x\in (e^{\lambda },e^{\mu })$ hence the values of $\lambda ,\mu$ are
$\lambda =-{\displaystyle \frac{\pi }{4}}$ and $\mu ={\displaystyle \frac{3\pi }{4}}$
Now the value of $-500\cos (\mu -\lambda )$ is
$\begin{array}{rl}& -500\cos (\mu -\lambda )=-500\cos ({\displaystyle \frac{3\pi }{4}}-(-{\displaystyle \frac{\pi }{4}}))\\ & =-500\cos ({\displaystyle \frac{3\pi }{4}}+{\displaystyle \frac{\pi }{4}})\\ & =-500\cos \pi \\ & =-500(-1)\\ & =500\end{array}$

Note: Please take the limits of $x$ for $\sin x>0$ as $0<x<\pi$. We have other ranges also for $x$ but it is the basic to consider the range from $0<x<\pi$. Be aware of the operations that we follow to simplify the range. The derivative of $\log x$ is ${\displaystyle \frac{1}{x}}$, mathematically ${\displaystyle \frac{d}{dx}}(\log x)={\displaystyle \frac{1}{x}}$ and the differentiation of the functions like $f\left(g\left(x\right)\right)$ is given by ${\displaystyle \frac{d}{dx}}\left(f\left(g\left(x\right)\right)\right)=f^{{}^{\prime }}\left(g\left(x\right)\right).g^{{}^{\prime }}\left(x\right)$