If f(x) = frac{(tan 1^{circ}) x + log_{e}(123 | vedclass

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(B) Let $f(x) = \frac{Ax + B}{Cx - A}$,where $A = 	an 1^{\circ}$,$B = \log_{e}(123)$,and $C = \log_{e}(1234)$.First,we calculate $f(f(x))$:$f(f(x)) =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $f(x) = \frac{Ax + B}{Cx - A}$,where $A = 	an 1^{\circ}$,$B = \log_{e}(123)$,and $C = \log_{e}(1234)$.First,we calculate $f(f(x))$:$f(f(x)) = \frac{A(\frac{Ax + B}{Cx - A}) + B}{C(\frac{Ax + B}{Cx - A}) - A} = \frac{A(Ax + B) + B(Cx - A)}{C(Ax + B) - A(Cx - A)} = \frac{A^2x + AB + BCx - AB}{ACx + BC - ACx + A^2} = \frac{x(A^2 + BC)}{A^2 + BC} = x$.Since $f(f(x)) = x$ for all $x$ in the domain,we have $f(f(x)) = x$ and $f(f(4/x)) = 4/x$.Therefore,$f(f(x)) + f(f(4/x)) = x + \frac{4}{x}$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,for $x > 0$:$x + \frac{4}{x} \geq 2 \sqrt{x \cdot \frac{4}{x}} = 2 \sqrt{4} = 4$.The least value is $4$.

If $f(x) = \frac{(\tan 1^{\circ}) x + \log_{e}(123)}{x \log_{e}(1234) - (\tan 1^{\circ})}$,$x > 0$,then the least value of $f(f(x)) + f(f(4/x))$ is $...........$.

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