MCQ
If $\text{x}=7+4\sqrt3$ and $xy = 1,$ then $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=$
- A$64$
- B$134$
- ✓$194$
- D$\frac{1}{49}$
✓
Answer
Correct option: C.
$194$
$\text{x}=7+4\sqrt3,\ \text{xy}=1\Rightarrow\text{y}=\frac{1}{\text{x}}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}\\ \ =\frac{7-4\sqrt3}{(7)^2-\big(4\sqrt3\big)^2}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3$
Now, $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=\frac{\text{y}^2+\text{x}^2}{\text{x}^2\text{y}^2}=\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}$
$\text{x}^2=\big(7+4\sqrt3\big)^2=49+48+56\sqrt3=97+56\sqrt3$
$\text{y}^2=\big(7-4\sqrt3\big)^2=49+48-56\sqrt3=97-56\sqrt3$
$\therefore\text{x}^2+\text{y}^2=97+56\sqrt3+97-56\sqrt3=194$
$\text{xy} = 1$
$\therefore\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}=\frac{194}{(1)^2}=194$
Hence, correct option is $(c).$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}$
$\therefore\text{y}=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}\\ \ =\frac{7-4\sqrt3}{(7)^2-\big(4\sqrt3\big)^2}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3$
Now, $\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}=\frac{\text{y}^2+\text{x}^2}{\text{x}^2\text{y}^2}=\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}$
$\text{x}^2=\big(7+4\sqrt3\big)^2=49+48+56\sqrt3=97+56\sqrt3$
$\text{y}^2=\big(7-4\sqrt3\big)^2=49+48-56\sqrt3=97-56\sqrt3$
$\therefore\text{x}^2+\text{y}^2=97+56\sqrt3+97-56\sqrt3=194$
$\text{xy} = 1$
$\therefore\frac{\text{x}^2+\text{y}^2}{(\text{xy})^2}=\frac{194}{(1)^2}=194$
Hence, correct option is $(c).$
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