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In the IEEE floating point representation the hexadecimal value $0\text{x}00000000$ corresponds to

1. The normalized value $2^{-127}$
2. The normalized value $2^{-126}$
3. The normalized value $+0$
4. The special value $+0$

what is the difference between Option A and D
The prefix ‘0x’ shows it’s a hexadecimal constant, in case anyone is unaware of 0x usage.

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$$\begin{array}{|c|c|c|}\hline \textbf{S} & \textbf{BE} & \textbf{M} & \textbf{Value} \\\hline 0/1 & {\color{Magenta} {\text{All 0’s}}} & {\color{Magenta} {\text{All 0’s}}} & 0 \\\hline 0 & {\color{Blue} {\text{All 1’s}}} & {\color{Magenta} {\text{All 0’s}}} & +\infty \\\hline 1 & {\color{Blue} {\text{All 1’s}}} & {\color{Magenta} {\text{All 0’s}}} & -\infty \\\hline 0/1 & {\color{Blue} {\text{All 1’s}}} & \text{Non-zero}& \text{NaN} \\\hline\end{array}$$The answer is option D.

Sir, what is the difference between option C and option D?
earlier to IEEE 758 representation, floating numbers couldn't differentiate -0 and +0. So, they come up with IEEE 758 floating poing representation in 32 bit and 64 bit and they marked +0 and -0 representation special that's all I read somewhere, don't know exact history.

D. The special value +0.

by

sir why not A??
this 32 bit ieee representation also correspond to normalize 2^(-127)
using bias=(2^(8-1))-1  value= (-1)^0 * 1.(23 0's) * 2^(0-bias)=1*2^(-127)
1 implicit here and not stored.  @arjun sir plz reply
srry got the point, we can't take all zero in exponent part in case of ieee 754 single precision, min value that can be stored is 2^(-126), which in case of when exponent field contain 1 in LSB position and all zero in mantissa part, so given format reserved for special value/ thanks
Answer is not A because Exponent and Mantisa both are zero which is equal to special value +0.
Very Nice and useful link it clears all IEEE Floating Point Representation Doubts.

Thank you Sir!.
Answer is not A because range of single precision normalized form is $\pm 2^{-126}$ to $\left (2 -2^{(-23)} \right ) \times 2 ^{127}$