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...
BODMAS=Bracket, Order, Division, Multiplication, Addition and Subtraction. In certain regions, PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition and Subtraction) is the synonym of BODMAS. It explains the order of operations to solve an expression.
- Statement; true
- Statement; false
- Statement; true
- Interrogative sentence; not a statement
- Imperative sentence; not a statement
- Exclamatory sentence; not a statement
Program to input any number and display number of odd numbers in it;
#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>
int main() {
int i, number, num1, num2=0, num3, num4 =0, rem, rem1, rem2, rev = 0;
printf("Enter your number ==> ");
scanf("%d", &number);
num1 = number;
while(num1 != 0) {
rem = num1%10;
rem1 = rem%2;
if (rem1 != 0) {
num2 = num2*10 + rem;
}
num1 /= 10;
}
num3 = num2;
while (num2 != 0) {
...
