Quantitative Aptitude Ques 919
Question: $ x+\frac{1}{2x}=2, $ then find the value of $ 8x^{2}+\frac{1}{x^{3}}. $
Options:
A) 48
B) 88
C) 40
D) 44
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, $ x+\frac{1}{2x}=2 $
On multiplying by 2 both sides, we get
$ 2x+\frac{2}{2x}=4 $
$ \Rightarrow $ $ 2x+\frac{1}{x}=4 $ On cubing both sides, we get $ {{( 2x+\frac{1}{x} )}^{3}}=4^{3} $
$ \Rightarrow $ $ {{(2x)}^{3}}+{{( \frac{1}{x} )}^{3}}+3\times 2x\times \frac{1}{x}( 2x+\frac{1}{x} )=64 $ $ [\because {{(a+b)}^{3}}=a^{3}+b^{3}+3ab(a+b)] $
$ \Rightarrow $ $ 8x^{2}+\frac{1}{x^{3}}+3\times 2x\times \frac{1}{x}( 2x+\frac{1}{x} )=64 $
$ \Rightarrow $ $ 8x^{3}+\frac{1}{x^{3}}+6\times 4=64 $
$ \Rightarrow $ $ 8x^{3}+\frac{1}{x^{3}}=64-24=40 $
BETA
