Given:
\( f(x) = \frac{(\tan 1)x + \log_{12}3}{x \log_{12}34 - \tan 1} \)
Let:
\( A = \tan 1, \quad B = \log_{12}3, \quad C = \log_{12}34 \)
Rewriting \(f(x)\):
\( f(x) = \frac{Ax + B}{xC - A} \)
Step 1: Evaluate \(f(f(x))\):
\( f(f(x)) = f \left( \frac{Ax+B}{xC-A} \right) \)
Substitute \(f(x)\) into \(f\):
\( f(f(x)) = \frac{A \left( \frac{Ax+B}{xC-A} \right) + B}{C \left( \frac{Ax+B}{xC-A} \right) - A} \)
Simplify:
\( f(f(x)) = \frac{A(Ax+B) + B(xC - A)}{C(Ax+B) - A(xC - A)} \)
Expand numerator and denominator:
Numerator: \( A^2x + AB + xBC - AB = x(A^2 + BC) \)
Denominator: \( ACx + BC - ACx + A^2 = A^2 + BC \)
Thus:
\( f(f(x)) = \frac{x(A^2 + BC)}{A^2 + BC} = x \)
Step 2: Evaluate \(f(f(4/x))\):
\( f(f(4/x)) = \frac{4}{x} \)
Step 3: Evaluate \(f(f(x)) + f(f(4/x))\):
\( f(f(x)) + f(f(4/x)) = x + \frac{4}{x} \)
Step 4: Apply AM-GM Inequality: Using AM-GM:
\( \frac{x + \frac{4}{x}}{2} \geq \sqrt{x \cdot \frac{4}{x}} \)
\( x + \frac{4}{x} \geq 4 \)
Final Answer: \( x + \frac{4}{x} \geq 4 \)
Let \(S=\left\{0∈(0,\frac{π}{2}) : \sum^{9}_{m=1} \sec(θ+(m-1)\frac{π}{6})\sec(θ+\frac{mπ}{6}) = -\frac{8}{\sqrt3}\right\}\)
Then,
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)