Quantitative Aptitude Ques 1930
Question: If $ 2\sin[ \frac{(2x+1)\pi }{2} ]=x^{2}+\frac{1}{x^{2}}, $ then the value of $ ( x-\frac{1}{x} ) $ is
Options:
A) $ -1 $
B) $ 2 $
C) $ 1 $
D) $ 0 $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ x^{2}+\frac{1}{x^{2}}=2\sin ( \frac{(2x+1)\pi }{2} ) $
$ \Rightarrow $ $ {{( x-\frac{1}{x} )}^{2}}+2=2\sin ( \frac{(2x+1)\pi }{2} ) $ $ [\because ,a^{2}+b^{2}={{(a-b)}^{2}}+2ab] $
$ \therefore $ $ x-\frac{1}{x}=0 $ [ $ \sin \frac{(2x+1)\pi }{2}=1 $ for all integer values of $ x $ ]
BETA
