| Find the value of \[\frac{1}{a+1}-\frac{1}{b+1},\]when \[a=\sqrt{2}+1,\]\[b=\sqrt{2}-1\] [SSC (CGL) 2014] |
A) \[0\]
B) \[1\]
C) \[1-\sqrt{2}\]
D) \[\sqrt{2}-1\]
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 \[1-\sqrt{2}.\] |
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