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
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
Updated On: Oct 27, 2024
- 3
- 4
- 42
- None of Above
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The Correct Option is A
Approach Solution - 1
The correct option is (A): 3.
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Approach Solution -2
Given: f(3x + 2y, 2x - 5y) = 19x , f(x,2x)=27
that means \(3x+2y=\frac{2x-5y}{2}\)
\(6x+4y=2x-5y\)
\(4x=-9y\), \(y=\frac{-4x}{9}\)
Put the value of y
\(3x + 2y=>3x+2(\frac{-4x}{9})=>\frac{19}{9}x\)
\(2x - 5y=>2x-5(\frac{-4x}{9})=>\frac{38}{9}x\)
Now put both the value in equation 1
\(f(\frac{19}{9}x,\frac{39}{9}x)=19x\)
Let's take common \(\frac{19}{9}\) from both the side
\(f(x,2x)=9x\)
put the value of f(x,2x)=27
\(27=9x=>x=3\)
So, the correct option is (A): 3.
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