If vectors are arranged as trigonal planar and have equal magnitude, it is an ideal case of zero resultant.
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...
a)When silver nitrate comes in contact with skin, it is reduced to white color metallic silver.In most cases, a black stain appears due to formation of finely divided silver which has black color.
The black color is due to the decomposition of the Ag+ ion into Ag, silver. Silver nitrate will slowly decompose with the silver ion reverting to elemental silver. The microscopic particles of silver are so small that they absorb light instead of reflecting it, and so appear black, instead of silver....
Here, f: A—>B
f(x)= (x-1)/(x+2) ; x -2
A= {-1,0,1,2,3,4}
B= {-2,1,-1/2,0,1/2,1/4,2/5}
Range = {-2,-1/2,0,1/4,1/2,2/5}
As range is not equal to codomain so the given function is not bijective.
We can make it bijective by omitting {1} from set B
