Question

Suppose \(f(x,y)\) is a real-valued function such that \(f(3x+2y,2x-5y)=19x\), for all real numbers \(x\) and \(y\) . The value of x for which \(f(x,2x) = 27\) , is

• A. 3
• B. 4
• C. 42
• D. None of Above

Answer 1

The Correct Option is A

Given: 
\( f(3x + 2y,\ 2x - 5y) = 19x \)  
\( f(x,\ 2x) = 27 \) 

Assume that: 
\( 3x + 2y = a \), and \( 2x - 5y = b \) 
We are given that: 
\( f(a,\ b) = 19x \) and \( f(x,\ 2x) = 27 \) 

Now we match the arguments of \( f(a,\ b) \) with \( f(x,\ 2x) \): 
Let \[ a = \frac{19}{9}x,\quad b = \frac{38}{9}x \] 
Now confirm if this satisfies the transformation: 
From \( 3x + 2y = \frac{19}{9}x \), solve for \( y \): \[ 3x + 2y = \frac{19}{9}x \Rightarrow 2y = \frac{19}{9}x - 3x = \frac{19 - 27}{9}x = -\frac{8}{9}x \Rightarrow y = -\frac{4}{9}x \] 
Put this value of \( y \) into \( 2x - 5y \): \[ 2x - 5y = 2x - 5 \cdot \left( -\frac{4}{9}x \right) = 2x + \frac{20}{9}x = \frac{38}{9}x \] 
So, the function becomes: \[ f\left( \frac{19}{9}x,\ \frac{38}{9}x \right) = 19x \] Let’s define \( u = \frac{19}{9}x \Rightarrow x = \frac{9}{19}u \) 
Then: \[ f(u,\ 2u) = 19x = 19 \cdot \frac{9}{19}u = 9u \] 
But we are told: \[ f(x,\ 2x) = 27 \] and from the general form above, \( f(x,\ 2x) = 9x \), so: \[ 9x = 27 \Rightarrow x = 3 \] 
Therefore, the correct option is (A): 3.