1 >a>0 > b Quantity: 1\. value of \frac{(a+b)^{2}-a^{2}-b^{2}}{(a+b)^{2}-(a^{2}-b^{2})} = \frac{1}{2(ab^{3}+ab)}

Instructions

In the given questions, two quantities are given, one as Quantity I and another as Quantity II. You have to determine relationship between two quantities and choose the appropriate option.

a: If quantity I ≥ quantity II
b: If quantity I > quantity II
c: If quantity I < quantity II
d: If quantity I = quantity II or the relationship cannot be established from the information that is given
e: If quantity quantity II

Question 11

1 >a>0 > b
Quantity:
1. value of $$\frac{(a+b)^{2}-a^{2}-b^{2}}{(a+b)^{2}-(a^{2}-b^{2})}$$ = $$\frac{1}{2(ab^{3}+ab)}$$

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SBI PO 2018

1 >a>0 > bQuantity:1. value of $$\frac{(a+b)^{2}-a^{2}-b^{2}}{(a+b)^{2}-(a^{2}-b^{2})}$$ = $$\frac{1}{2(ab^{3}+ab)}$$

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1 >a>0 > b
Quantity:
1. value of $$\frac{(a+b)^{2}-a^{2}-b^{2}}{(a+b)^{2}-(a^{2}-b^{2})}$$ = $$\frac{1}{2(ab^{3}+ab)}$$