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

Last updated date: 24th Jun 2024
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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 $-500\cos \left( \mu -\lambda \right)$

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

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

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