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Question: If the function \(f\left( x \right)=\sin \left( \log x \right)-\cos \left( \log x \right)\) strictly...

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

Explanation

Solution

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 xx and we equate the result range with the given range to find the values of λ,μ\lambda ,\mu from that values we can find the required value of −500cos⁡(μ−λ)-500\cos \left( \mu -\lambda \right)

Complete step by step answer:
Given that, f(x)=sin⁡(log⁡x)−cos⁡(log⁡x)f\left( x \right)=\sin \left( \log x \right)-\cos \left( \log x \right) strictly increase in the interval (eλ,eμ)\left( {{e}^{\lambda }},{{e}^{\mu }} \right), then
f′(x) > 0{{f}^{'}}\left( x \right)\text{ }>\text{ }0
The value of f′(x){{f}^{'}}\left( x \right) is
f′(x)=ddx[sin⁡(log⁡x)−cos⁡(log⁡x)] =ddx[sin⁡(log⁡x)]−ddx[cos⁡(log⁡x)]\begin{aligned} & {{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{aligned}
Use the formulas ddx(sin⁡x)=cos⁡x\dfrac{d}{dx}\left( \sin x \right)=\cos x, ddx(cos⁡x)=−sin⁡x\dfrac{d}{dx}\left( \cos x \right)=-\sin x in the above equation, then
f′(x)=cos⁡(log⁡x).ddx(log⁡x)+sin⁡(log⁡x)ddx(log⁡x){{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 ddx(log⁡x)=1x\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}, then
f′(x)=cos⁡(log⁡x).1x+sin⁡(log⁡x).1x =1x[cos⁡(log⁡x)+sin⁡(log⁡x)]\begin{aligned} & {{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{aligned}
if the function f(x)f\left( x \right) strictly increase in the interval (eλ,eμ)\left( {{e}^{\lambda }},{{e}^{\mu }} \right), then
f′(x) > 0 1x[sin⁡(log⁡x)+cos⁡(log⁡x)] > 0 sin⁡(log⁡x)+cos⁡(log⁡x) > 0\begin{aligned} & {{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{aligned}
Multiply and divide by 2\sqrt{2} in the above expression, then we have
2(12sin⁡(log⁡x)+12cos⁡(log⁡x)) > 0\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⁡π4=cos⁡π4=12\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}in the above expression, then
cos⁡π4.sin⁡(log⁡x)+sin⁡π4cos⁡(log⁡x) > 0\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⁡(x+y)\sin x.\cos y+\cos x.\sin y=\sin \left( x+y \right) in the above expression, then we have
sin⁡(π4+log⁡x) > 0\sin \left( \dfrac{\pi }{4}+\log x \right)\text{ }>\text{ }0
The graph of y=sin⁡xy=\sin x is given below

Form the above equation we have value of sin⁡x > 0\sin x\text{ }>\text{ }0 for 0 < x < π0\text{ }<\text{ }x\text{ }<\text{ }\pi , so the value of sin⁡(π4+log⁡x) > 0\sin \left( \dfrac{\pi }{4}+\log x \right)\text{ }>\text{ }0
For
0 < π4+log⁡x < π −π4 < log⁡x < π−π4 −π4 < log⁡x < 3π4 e−π4 < x < e3π4\begin{aligned} & 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{aligned}
∴\therefore x∈(e−π4,e3π4)x\in \left( {{e}^{-\dfrac{\pi }{4}}},{{e}^{\dfrac{3\pi }{4}}} \right)
But given that x∈(eλ,eμ)x\in \left( {{e}^{\lambda }},{{e}^{\mu }} \right) hence the values of λ,μ\lambda ,\mu are
λ=−π4\lambda =-\dfrac{\pi }{4} and μ=3π4\mu =\dfrac{3\pi }{4}
Now the value of −500cos⁡(μ−λ)-500\cos \left( \mu -\lambda \right) is
−500cos⁡(μ−λ)=−500cos⁡(3π4−(−π4)) =−500cos⁡(3π4+π4) =−500cos⁡π =−500(−1) =500\begin{aligned} & -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{aligned}

Note: Please take the limits of xx for sin⁡x > 0\sin x\text{ }>\text{ }0 as 0 < x < π0\text{ }<\text{ }x\text{ }<\text{ }\pi . We have other ranges also for xx but it is the basic to consider the range from 0 < x < π0\text{ }<\text{ }x\text{ }<\text{ }\pi . Be aware of the operations that we follow to simplify the range. The derivative of log⁡x\log x is 1x\dfrac{1}{x}, mathematically ddx(log⁡x)=1x\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x} and the differentiation of the functions like f(g(x))f\left( g\left( x \right) \right) is given by ddx(f(g(x)))=f′(g(x)).g′(x)\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)