Question

if f(x)= \(\frac{(tan1\degree)x+\log_e(123)}{x\log_e(1234)-(tan1\degree)},x>0\) , then the least value of \(f(f(x))+f(f(\frac{4}{x}))\)is

• A. 0
• B. 2
• C. 4
• D. 8

Answer 1

The Correct Option is C

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: \( x + \frac{4}{x} \geq 4 \)