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If the sum of squares of two numbers is $97$, then which one of the following cannot be their product?

  1. $-32$
  2. $48$
  3. $64$
  4. $16$
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Let $x$ and $y$ be the two numbers.

$ x^{2} + y^{2} = 97 \quad \longrightarrow (1) $

The geometric mean cannot exceed the arithmetic mean.  $ \boxed{ \text{AM} \geqslant \text{GM}} $

$ \Rightarrow \boxed{\frac{a_{1} + a_{2}  + \dots + a_{n} } {n} \geqslant \sqrt[n]{a_{1} a_{2} \dots a_{n}}} $

Now$, \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{x^{2} \cdot y^{2}} $

$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{(xy)^{2}} $

$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant xy $

$ \Rightarrow x^{2} + y^{2} \geqslant 2xy $

$ \Rightarrow 97 \geqslant 2xy \quad [\because \text{From equation (1)}]$

$ \Rightarrow 2xy \leqslant 97 $

$ \Rightarrow xy \leqslant \frac{97}{2} $

$ \Rightarrow \boxed{xy \leqslant 48. 5}$

So$,xy$ cannot be more than $48. 5.$

$\therefore$ Only option $\text{(C)}$ not possible.

Correct Answer $: \text{C}$
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$\textrm{Let a,b are those number.}$

$\Rightarrow$ $a^2+b^2=97$

$\Rightarrow$ $a^2+b^2-2ab=97-2ab$  $\textrm{(adding -2ab in both side)}$

$\Rightarrow$ $(a-b)^2=97-2ab$

$\Rightarrow$ $\textrm{97-2ab $\geq$=0}$

$\Rightarrow$ $\textrm{ab$\leq48.5$}$

$\textrm{Hence ab $\neq$64,Option 3}$
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