If the minimum value of cos(sinh(log x) + cos | vedclass

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(D) Given $f(x) = \cos(\sinh(\log x) + \cosh(\log x))$.Using the definitions $\sinh(u) = \frac{e^u - e^{-u}}{2}$ and $\cosh(u) = \frac{e^u + e^{-u}}{2

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$D) 1$

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(D) Given $f(x) = \cos(\sinh(\log x) + \cosh(\log x))$.Using the definitions $\sinh(u) = \frac{e^u - e^{-u}}{2}$ and $\cosh(u) = \frac{e^u + e^{-u}}{2}$,we have:$\sinh(\log x) + \cosh(\log x) = \frac{e^{\log x} - e^{-\log x}}{2} + \frac{e^{\log x} + e^{-\log x}}{2} = \frac{2e^{\log x}}{2} = x$.Thus,$f(x) = \cos(x)$.The minimum value of $\cos(x)$ is $-1$.Therefore,$k = -1$.We need to find $\cosh(k+1) = \cosh(-1+1) = \cosh(0)$.Since $\cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1+1}{2} = 1$.

If the minimum value of $\cos(\sinh(\log x) + \cosh(\log x))$ is $k$,then $\cosh(k+1) =$

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