If species area relationships are analyzed for very large areas like…

The correct answer is Option (3) → 0.6 to 1.2

The species-area relationship says that species richness rises as the area explored rises. On a log scale it is a straight line:

    \( \displaystyle \log S = \log C + Z \log A \)

Here S is species richness, A is area, C is the Y-intercept and Z is the slope of the line. A larger Z means that richness rises faster as the area grows. For areas of ordinary size, Z is about 0.1 to 0.2 whatever the region or the group of organisms. Only for very large areas does it change.

(1) 0.1 to 0.2 — Incorrect. This is the usual range of Z for areas of ordinary size.

(2) 0.1 to 0.5 — Incorrect. This is not a range of Z for any case in the book.

(3) 0.6 to 1.2 — Correct. Over very large areas, such as whole continents, the line is much steeper. For fruit-eating birds and mammals in the tropical forests of different continents, Z is about 1.15.

(4) 1.3 to 1.6 — Incorrect. This is steeper than the range found even for continents.

NCERT Extract:
Patterns of Biodiversity – “… Ecologists have discovered that the value of Z lies in the range of 0.1 to 0.2, regardless of the taxonomic group or the region (whether it is the plants in Britain, birds in California or molluscs in New York state, the slopes of the regression line are amazingly similar). But, if you analyse the species-area relationships among very large areas like the entire continents, you will find that the slope of the line to be much steeper (Z values in the range of 0.6 to 1.2).”