MCQ
If $\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3,$ then:
- ✓$a = 2, b = 1$
- B$a = 2, b = -1$
- C$a = -2, b = 1$
- D$a = b = 1$
✓
Answer
Correct option: A.
$a = 2, b = 1$
$\frac{\sqrt3-1}{\sqrt3+1}$
Multiplying and dividing by the rationalisation factor of denominator, we get
$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}$
$=\frac{\big(\sqrt3-1\big)^2}{\big(\sqrt3\big)^2-1^2}$
$=\frac{3-2\sqrt3+1}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=\frac{2(2-\sqrt3)}{2}$
$=2-\sqrt3$
Comparing with $\text{a}-\text{b}\sqrt3,$ we get $a = 2$ and $b = 1.$
Hence, correct option is $(a).$
Multiplying and dividing by the rationalisation factor of denominator, we get
$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}$
$=\frac{\big(\sqrt3-1\big)^2}{\big(\sqrt3\big)^2-1^2}$
$=\frac{3-2\sqrt3+1}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=\frac{2(2-\sqrt3)}{2}$
$=2-\sqrt3$
Comparing with $\text{a}-\text{b}\sqrt3,$ we get $a = 2$ and $b = 1.$
Hence, correct option is $(a).$
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