SATHEE: Quantitative Aptitude Ques 1321

Quantitative Aptitude Ques 1321

Question: If $ x=\frac{\sqrt{2}+1}{\sqrt{2}-1} $ and $ y=\frac{\sqrt{2}-1}{\sqrt{2}+1}, $ then the value of $ (x^{2}+y^{2}) $ is

Options:

A) 34

B) 36

C) 32

D) 38

Show Answer

Answer:

Correct Answer: A

Solution:

Since, $ x=\frac{\sqrt{2}+1}{\sqrt{2}-1} $ and $ y=\frac{\sqrt{2}-1}{\sqrt{2}+1} $

$ \therefore $ $ x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{{{(\sqrt{2}+1)}^{2}}}{2-1} $ $ =\frac{2+1+2\sqrt{2}}{1}=3+2\sqrt{2} $ Similarly, $ y=\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1} $ $ =\frac{2+1-2\sqrt{2}}{2-1}=\frac{3-2\sqrt{2}}{1} $ Now, $ (x^{2}+y^{2})={{(3+2\sqrt{2})}^{2}}+{{(3-2\sqrt{2})}^{2}} $ $ =9+8+2\times 3\times 2\sqrt{2} $ $ +9+8-2\times 3\times 2\sqrt{2} $ $ =18+16=34 $

Quantitative Aptitude Ques 1320

Quantitative Aptitude Ques 1322

Quantitative Aptitude Ques 1321

Question: If $ x=\frac{\sqrt{2}+1}{\sqrt{2}-1} $ and $ y=\frac{\sqrt{2}-1}{\sqrt{2}+1}, $ then the value of $ (x^{2}+y^{2}) $ is

Options:

Answer:

Solution:

Bank Preparation

Bank Courses

NCERT Books

Important Resources

Forum