A) \[{{a}^{2}}+{{b}^{2}}\]
B) \[{{a}^{2}}-{{b}^{2}}\]
C) \[{{a}^{2}}+2{{b}^{2}}\]
D) None of these
Correct Answer: A
Solution :
| Let \[t=x+\sqrt{{{x}^{2}}+{{b}^{2}}}\] |
| \[\Rightarrow \]\[\frac{1}{t}=\frac{1}{x+\sqrt{{{x}^{2}}+{{b}^{2}}}}=\frac{\sqrt{{{x}^{2}}+{{b}^{2}}}-x}{{{b}^{2}}}\] |
| \[\Rightarrow \]\[t-\frac{{{b}^{2}}}{t}=2x\And t+\frac{{{b}^{2}}}{t}=2\sqrt{{{x}^{2}}+{{b}^{2}}}\] |
| \[\therefore \] \[2(a-x)(x+\sqrt{{{x}^{2}}+{{b}^{2}}})\] |
| \[=\left( 2a-t+\frac{{{b}^{2}}}{t} \right)\left( t \right)=2at-{{t}^{2}}+{{b}^{2}}\] |
| \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}-({{a}^{2}}-2at+{{t}^{2}})\] |
| \[={{a}^{2}}+{{b}^{2}}-{{(a-t)}^{2}}\le {{a}^{2}}+{{b}^{2}}\] |
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