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The interrupt vector table is always created in the first 1K area of the memory. Justify the statement.
While the CPU receives an interrupt type number start from the PIC, this uses this number to look up the corresponding interrupt vector into memory. Here are 256 interrupt types. All interrupt vectors occupies 4 bytes. Thus, a total of 4 x 256 = 1K bytes of memory are reserved at the starting of the processor memory address space for storing interrupt vectors.
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#z =x1+2.5x2, with constraints, x1+x2>=2, x2 x1+x2 x1=3.question..
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explain with figure
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