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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II)Cr2+ is reducing agent as its configuration changes from d4 to d3, when it is oxidized to Cr3+ .The d3 configuration have a half-filled t2g level which is very stable. On the other hand, the reduction of Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability hence Mn3+ acts as oxidizing agent.
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#include<stdio.h>
#include<conio.h>
#include<string.h>
int main() {
int i, number, num1, num2=0;
char str1[50];
num1 = number;
for(i=2; i<=20; i+=2) {
printf("%d", i);
printf(", ");
}
return 0;
}
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