Since T is divisible by 12. Let's assume T to be 12. Then, \(T^2\) =144.
Then the minimum value for which \(T^2\) is not an integer when divided by 2^a is when a=5.
Hence D is the answer.

Explanation

If t is divisible by 12, then \(t^2\) must be divisible by 144 or \(2 \times 2 \times 2 \times 2 \times 3 \times 3\). Therefore, \(t^2\) can be divided evenly by 2 at least four times, so a must be at least 5 before \(\frac{t^2}{2^a}\) might not be an integer.
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If t is divisible by 12 and we need to find the least possible integer value of a for which t^2/2^a might not be an integer
So, we need to take the smallest possible positive value of t
[positive as 0 is divisible by all numbers so it will be divisible by all values of 2^a]
=> t = 12 = \(2^2 * 3\)
Now, \(t^2\) = (\(2^2 * 3\))^2 = \(2^4 * 3^2\)
So min value of a for which \(t^2\) might not be an integer will be 5

So, answer will be D
Hope it helps!
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You mean D right ?

Thanks for pointing it out. Edited the post accordingly.


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