Question : If $\left \{ \frac{1}{2}(a-b) \right \}^{2}+ab=p(a+b)^{2}$, the value of $p$ is:
Option 1: $p=4$
Option 2: $p=\frac{1}{2}$
Option 3: $p=\frac{1}{4}$
Option 4: $p=2$
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Correct Answer: $p=\frac{1}{4}$
Solution :
Given: $\left \{ \frac{1}{2}(a–b) \right \}^{2}+ab=p(a+b)^{2}$
⇒ $\frac{1}{4}(a - b)^2 + ab = p(a + b)^2$
Multiplying 4 on both sides,
⇒ $4 × \frac{1}{4}(a - b)^2 + 4ab = 4p(a + b)^2$
⇒ $(a - b)^2 + 4ab = 4p(a + b)^2$
⇒ $a^2 + b^2 - 2ab + 4ab = 4p(a + b)^2$
⇒ $a^2 + b^2 + 2ab = 4p(a + b)^2$
⇒ $(a + b)^2 = 4p(a + b)^2$
⇒ $4p = 1$
⇒ $p = \frac{1}{4}$
Hence, the correct answer is $p = \frac{1}{4}$.
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