Question: Which of the following equations are equivalent?
I. $ {{( \frac{1}{2}M+\frac{2}{3}N )}^{2}} $
II. $ \frac{4}{9}N^{2}+\frac{1}{4}M^{2}+\frac{2}{3}MN $
III. $ ( \frac{M}{2}+\frac{2}{3}N )( \frac{1}{2}M-\frac{2}{3}N ) $
IV. $ \frac{1}{4}{{( M+\frac{4}{3}N )}^{2}} $
Options:
A) II and III
B) I and IV
C) I and II
D) I and III
E) I, II and IV
Show Answer
Answer:
Correct Answer: E
Solution:
- Simplifying .all the equations,
I. $ {{( \frac{1}{2}M+\frac{2}{3}N )}^{2}}=\frac{1}{4}M^{2}+\frac{4}{9}N^{2}+\frac{2}{3}MN $
II. $ \frac{4}{9}N^{2}+\frac{1}{4}M^{2}+\frac{2}{3}MN=\frac{1}{4}M^{2}+\frac{4}{9}N^{2}+\frac{2}{3}MN $
III. $ ( \frac{M}{2}+\frac{2}{3}N )( \frac{1}{2}M-\frac{2}{3}N ) $
$ =\frac{1}{4}M^{2}+\frac{1}{3}MN-\frac{1}{3}MN-\frac{4}{9}N^{2}=\frac{1}{4}M^{2}-\frac{4}{9}N^{2} $
IV. $ \frac{1}{4}{{( M+\frac{4}{3}N )}^{2}}=\frac{1}{4}[ M^{2}+\frac{16}{9}N^{2}+\frac{8}{3}MN ] $
$ =\frac{1}{4}M^{2}+\frac{4}{9}N^{2}+\frac{2}{3}MN $
From the above four solutions, we find that I, II and IV are equivalent.
BETA
