Question

If \(f(x)\) is a linear polynomial such that \[ f(ax+by)=af(x)+bf(y) \] for all \(x,y\in\mathbb{R}\). If \(f(0)=7\) and \(f'(0)=5\), then \[ \frac{f(1)+f(-1)}{a+b}= \]

• A. 10
• B. 12
• C. 2
• D. 14

Hint:

For a linear polynomial, \[ f(x)=mx+c, \] \[ \boxed{m=f'(x)=f'(0),\qquad c=f(0).} \] In functional equations, substitute the polynomial directly and compare coefficients or constant terms to obtain unknown parameters.

Answer 1

The Correct Option is D

Step 1: Write the linear polynomial. Since \(f(x)\) is a linear polynomial, \[ f(x)=mx+c. \] Given, \[ f(0)=7, \] we get \[ c=7. \] Also, \[ f'(x)=m. \] Since \[ f'(0)=5, \] we obtain \[ m=5. \] Hence, \[ \boxed{f(x)=5x+7.} \]

Step 2: Use the given functional equation to find \(a+b\). Given, \[ f(ax+by)=af(x)+bf(y). \] Substituting \(f(x)=5x+7\), \[ 5(ax+by)+7 = a(5x+7)+b(5y+7). \] Simplifying, \[ 5ax+5by+7 = 5ax+7a+5by+7b. \] Therefore, \[ 7=7(a+b), \] which gives \[ \boxed{a+b=1.} \]

Step 3: Evaluate the required expression. Now, \[ f(1)=5(1)+7=12, \] and \[ f(-1)=5(-1)+7=2. \] Hence, \[ \frac{f(1)+f(-1)}{a+b} = \frac{12+2}{1} = 14. \] Thus, \[ \boxed{14} \] is the correct answer.