MCQ
If $\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$ and $\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3},$ then $\text{x}+\text{y}+\text{xy}=$
- ✓$9$
- B$5$
- C$17$
- D$7$
✓
Answer
Correct option: A.
$9$
$\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$
$\therefore\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\\ \ =\frac{\big(\sqrt5+\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$
$\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
$\therefore\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}\\ \ =\frac{\big(\sqrt5-\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}$
$\text{xy}=\big(4+\sqrt{15}\big)\big(4-\sqrt{15}\big)=16-15=1$
Now, $\text{x}+\text{y}+\text{xy}=4+\sqrt{15}+4-\sqrt{15}+1\\ \ =4+4+1=9$
Hence, correct option is $(a).$
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