(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) 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 $\cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1+1}{2} = 1$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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

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A
$A) \frac{e+e^{-1}}{2}$
B
$B) \frac{e^2+e^{-2}}{2}$
C
$C) e$
D
$D) 1$

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Trigonometrical Ratios, Functions and Identities

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Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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