Question

If \(3f(x+2)+4f\left(\dfrac{1}{x+2}\right)=4x\), \(x\neq-2\), then \(f(4)\) is:

• A. 7
• B. 52/7
• C. 8
• D. None of the above

Hint:

Replace x with a value that turns 1/(x+2) into the same argument the other term used, so you get a second equation in the same two unknowns.

Answer 1

The Correct Option is D

Step 1: Substitute to simplify the argument.
Let \(y=x+2\), so \(x=y-2\). The equation \(3f(x+2)+4f\left(\dfrac{1}{x+2}\right)=4x\) becomes \[ 3f(yy)+4f\left(\dfrac{1}{y}\right) = 4(y-2) = 4y-8 \qquad (ii) \]

Step 2: Build a second equation by swapping the roles.
Equation (ii) links \(f(y)\) and \(f\left(\dfrac{1}{y}\right)\), but with only one equation we cannot isolate \(f(y)\) yet. To get a second, independent equation, replace \(x\) by the value that makes \(\dfrac{1}{x+2}\) equal to \(y\). Setting \(\dfrac{1}{x+2}=y\) gives \(x=\dfrac{1-2y}{y}\). Feeding this into the original relation swaps which term becomes \(f(y)\) and which becomes \(f\left(\dfrac{1}{y}\right)\): \[ 3f\left(\dfrac{1}{y}\right)+4f(yy) = 4\left(\dfrac{1-2y}{y}\right) \qquad (ii) \]

Step 3: Eliminate \(f\left(\dfrac{1}{y}\right)\) between (ii) and (ii).
Multiply (ii) by \(3\) and (ii) by \(4\), so the coefficient of \(f\left(\dfrac{1}{y}\right)\) becomes \(12\) in both: \[ 9f(yy)+12f\left(\dfrac{1}{y}\right) = 12(y-2) \] \[ 16f(yy)+12f\left(\dfrac{1}{y}\right) = 16\left(\dfrac{1-2y}{y}\right) \] Subtracting the first from the second: \[ 7f(yy) = 16\left(\dfrac{1-2y}{y}\right) - 12(y-2) \]

Step 4: Solve for f(yy) and plug in y=4. \[ f(yy) = \dfrac{1}{7}\left[16\left(\dfrac{1-2y}{y}\right)-12(y-2)\right] \] At \(y=4\): \(\dfrac{1-2(4)}{4}=-\dfrac{7}{4}\), so \[ f(44) = \dfrac{1}{7}\left[16\times\left(-\dfrac{7}{4}\right) - 12(4-2)\right] = \dfrac{1}{7}\left[-28-24\right] = -\dfrac{52}{7} \]

Step 5: Compare with the given options.
\(f(4)=-\dfrac{52}{7}\) is negative, but none of the numeric options 7, 52/7, or 8 match \(-\dfrac{52}{7}\), so the correct choice is the last listed option, 'None of the above'.

Final Answer:
\(f(4)=-\dfrac{52}{7}\), which is not listed among the numeric options. \[ \boxed{-\dfrac{52}{7}} \]