Here, f(x)=x2 -6 logx-3=0
f(2)=4-6 log2-3=-0.806
f(3)=9-6 log3-3=3.1373
f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.
Hence, the root lies between 2 and 3
c0 =(2+3)/2=2.5
f(2.5)=6.25-6 log 2.5-3=0.8623
Now
| n | a(-ve) | b(+ve) | cn | f(cn) |
| 0 | 2 | 3 | 2.5 | 0.8623 |
| 1 | 2 | 2.5 | 2.25 | -0.050595 |
| 2 | 2.25 | 2.5 | 2.375 | 0.38664 |
| 3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |
| 4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |
| 5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |
From the table,
f(2.265625)=0.001928<10-2
Therefore, the...
The answer is A
In the depletion layer, electrons combine with holes producing charge density so that region is depleted of charge carriers but it has charge density .
Conclusion: Neutral but still having non zero charge density .
In mitosis, the prophase is further understood by dividing it into the given sub-stages:
1.Leptotene :
In this stage the nucleus enlarges in size in the chromosome. The chromosomes appear thin, thread-like and single-stranded in this stage. They have swollen or beaded structures along their length and their ends appear converged towards one side of the nucleus called bouquet.
2.Zygotene:
In this stage, the identical chromosomes come together and form bivalent or homologous pairs. Further, ...
