Quantitative Aptitude Ques 1270
Question: Find the value of $ \frac{1}{a+1}-\frac{1}{b+1}, $ when $ a=\sqrt{2}+1, $ $ b=\sqrt{2}-1 $
Options:
A) $ 0 $
B) $ 1 $
C) $ 1-\sqrt{2} $
D) $ \sqrt{2}-1 $
Show Answer
Answer:
Correct Answer: C
Solution:
- Given expression be $ =\frac{1}{a+1}-\frac{1}{b+1} $ Where, $ a=\sqrt{2}+1 $ and $ b=\sqrt{2}-1 $ Taking first part, $ \frac{1}{a+1}=\frac{1}{\sqrt{2}+1+1}=\frac{1}{\sqrt{2}+2} $ $ =\frac{1}{\sqrt{2}+2}\times \frac{\sqrt{2}-2}{\sqrt{2}-2} $ $ =\frac{\sqrt{2}-2}{{{(\sqrt{2})}^{2}}-{{(2)}^{2}}}=\frac{2-\sqrt{2}}{2} $ (i) Now, taking second part $ \frac{1}{b+1}=\frac{1}{\sqrt{2}-1+1}=\frac{1}{\sqrt{2}} $ $ =\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} $ … (ii) Putting Eqs. (i) and (ii) in the given expression, we get $ \frac{1}{a+1}-\frac{1}{b+1}=\frac{2-\sqrt{2}}{2}-\frac{\sqrt{2}}{2} $ $ =\frac{2-\sqrt{2}}{2}=\frac{2-2\sqrt{2}}{2}=1-\sqrt{2} $
$ \therefore $ Value of expression is $ 1-\sqrt{2}. $
BETA
