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# Evaluate ${\left( {\dfrac{a}{{2b}} + \dfrac{{2b}}{a}} \right)^2} - {\left( {\dfrac{a}{{2b}} - \dfrac{{2b}}{a}} \right)^2} - 4$A) 2B) abC) aD) 0 Verified
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Hint: Students if we observe the problem, it is somewhat like ${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} - 4$. This is the hint to solve the question. Instead of solving the original question we will consider the terms as a and b. and then solve it using the identity. And the answer of the identity will be replaced by the considered values.

Complete step by step solution:
Given that,
${\left( {\dfrac{a}{{2b}} + \dfrac{{2b}}{a}} \right)^2} - {\left( {\dfrac{a}{{2b}} - \dfrac{{2b}}{a}} \right)^2} - 4$
Now if we compare it is same as ${\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} - 4$
We know that,
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
So let a=$\dfrac{a}{{2b}}$ and b=$\dfrac{{2b}}{a}$ . now we will expand the bracket or equation we substituted.
$= {\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} - 4$
Taking the expansion, we get,
$= {a^2} + 2ab + {b^2} - \left( {{a^2} - 2ab + {b^2}} \right) - 4$
Multiplying the terms by the minus sign we get,
$= {a^2} + 2ab + {b^2} - {a^2} + 2ab - {b^2} - 4$
Now cancel the same terms but with different signs,
$= 2ab + 2ab - 4$
$= 4ab - 4$
$= 4 \times \dfrac{a}{{2b}} \times \dfrac{{2b}}{a} - 4$
$= 4 - 4$
$= 0$
Therefore, ${\left( {\dfrac{a}{{2b}} + \dfrac{{2b}}{a}} \right)^2} - {\left( {\dfrac{a}{{2b}} - \dfrac{{2b}}{a}} \right)^2} - 4 = 0$. So, option (D) is the correct answer.