darlingriyarai wrote:
A function f is defined by f(x,y) = x^2 - y. Is the value of f(3,b) less than the value of f(a,3)?
(1) a < b
(2) a + 4 < 0
OA: C\(f(x,y) = x^2 - y\)
\(f(3,b) = 9 -b\)
\(f(a,3) = a^2-3\)
Question Stem : \(Is \quad f(a,3)>f(3,b)?\)
\(Is \quad a^2-3>9 -b? \quad or \quad Is \quad a^2 +b>12?\)
\((1) \quad a < b\)
Taking \(a = 1\) and \(b = 2\) , \(1<2\)
\(Is \quad a^2 +b>12?\quad No\) , as \(1^2+2 =3\) is less than \(12\).
Taking \(a = 2\) and \(b = 10\) , \(2<10\)
\(Is \quad a^2 +b>12?\quad Yes\) , as \(2^2+10 =14\) is greater than \(12\).
Statement \(1\) alone is not sufficient
\((2) \quad a + 4 < 0 \quad i.e \quad a<-4\)
Value of \(b\) can be anything so Statement \(2\) alone is not sufficient.
Combining \((1)\) and \((2)\) , we get \(a<-4\) and \(b≥-4\)
The minimum value of \(a^2\) would be slightly more than \(16\).
The minimum value of \(b\) would be \(-4\)
So Minimum value of \(a^2+b\) would be \(16.X-4 =12.X\) where \(X\) is greater than \(0\),So \(a^2+b\) would be always greater than \(12\).
Combining \((1)\) and \((2)\) \(Is \quad a^2 +b>12?\quad Yes\)