Question

$let \: f(x) = \sqrt{1 + {x}^{2} } \: then$
1) f(xy) = f(x).f(y)
2) f(xy) ≥ f(x).f(y)
3) f(xy) ≤ f(x).f(y)
4) None of these

Choose the appropriate answer with explanation.​

Answer 1

Answer: (iii) i.e., f(xy) < f(x).f(y)

$f(xy) = \sqrt{1 + {x}^{2} {y}^{2} } \\ {(f(xy)})^{2} = (1) + ( {x}^{2} {y}^{2}) \\ f(x) = \sqrt{1 + {x}^{2} } > > (i) \\ f(y) = \sqrt{1 + {y}^{2} } > > (ii) \\ \\ multiplying \: (i) \: and \: (ii) \: we \: get$

$f(x).f(y) = ( \sqrt{1 + {x}^{2} } )( \sqrt{1 + {y}^{2} } ) \\ = \sqrt{(1 + {x}^{2} )(1 + {y}^{2} )} \\ = \sqrt{1 + {x}^{2} + {y}^{2} + {x}^{2} {y}^{2} } \\ = \sqrt{ (1 + {x}^{2} + {y}^{2} ) + ( {x}^{2} {y}^{2} )} \\ {(f(x).f(y)})^{2} = (1 + {x}^{2} + {y}^{2} ) + ( {x}^{2} {y}^{2} )$

Now it can be observed that, [f(x).f(y)]² > [f(xy)]²

Therefore, it can be concluded that f(x).f(y) > f(xy).

Thus, option 3.