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
Yes, a vector which has zero magnitude is also a vector in case of two vectors travelling in opposite directions with equal magnitudes. At this case, the resultant vector has zero magnitude but it is still a vector. We call it a null vector.
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...
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