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