Question 66

If $$[\frac{1}{2}(a-b)]^2+ab=p(a+b)^2$$, then the value of p is

Solution

Expression : $$[\frac{1}{2}(a-b)]^2+ab=p(a+b)^2$$

=> $$\frac{a^2+b^2-2ab}{4}+ab=p(a+b)^2$$

=> $$\frac{a^2+b^2-2ab+4ab}{4}=p(a+b)^2$$

=> $$a^2+b^2+2ab=4p(a+b)^2$$

=> $$(a+b)^2=4p(a+b)^2$$

=> $$4p=1$$

=> $$p=\frac{1}{4}$$

=> Ans - (C)

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