Question : The value of $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$ lies between:
Option 1: 3 and 3.5
Option 2: 2 and 2.5
Option 3: 1.5 and 2
Option 4: 2.5 and 3
Correct Answer: 3 and 3.5
Solution :
Given: $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$
= $\frac{(7+3 \sqrt{5})(3-\sqrt{5})-(7-3\sqrt5)(3+\sqrt{5})}{(3)^2-(\sqrt{5})^2}$
= $\frac{21+9\sqrt{5}-7\sqrt5-15+21+7\sqrt5-9\sqrt5-15}{9-5}$
= $\frac{6+6}{4}$
= $\frac{12}{4}$
= 3
Thus, 3 lies between 3 and 3.5
Hence, the correct answer is 3 and 3.5
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