Sunday, December 17, 2017

CSITauthority News: December 17, 2017 at 02:02PM

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In recent: Frustrated 8thSem CSIT students calling Dean's Office, IOST, by the minute, to ask when their 6thSem result is going to be published 😂😂😂 ps. In all seriousness, aaja voli samma aucha vanera reply aayeko cha. Tara last week samma Sunday pani vaniyeko thyo. Aaja aaos hopefully.. #fingerscrossed



News mirrored from: CSIT Authority
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Wednesday, August 2, 2017

Entrance form open for B.Sc. CSIT Admission 2074

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Tribhuvan University, Institute of Science and Technology (IOST) announces entrance examination notice for the enrollment in Bachelor of Computer Science and Information Technology (B.Sc. CSIT program) for the academic year 2074 in its constituent and affiliated colleges.
  • Application form fee: Rs. 1200
  • Last date to submit Application form: 2074 Bhadra 14
  • Last date to submit Application form(with Double fee): 2074 Bhadra 19
  • Entrance exam date: 2074 Bhadra 24 ( 12AM to 2PM)
BONUS NEWS!! ST. XAVIER'S COLLEGE ENTRANCE EXAM IS GOING TO BE HELD ON 10th September 2017. This college conducts an additional internal exam. Please hurry up to get your entrance examination form if you are interested.

Currently 54 Colleges offers B.Sc. CSIT program in Nepal with affiliation from Tribhuvan University. For details, eligibility, requirements, support and so on regarding Admission and the process, we highly recommend you to visit our Admission Page.
Official Notice by TU, IOST

TU Admission Form 2074 :

TU Form:
TU IOST Admission Form 2074

Here are some MODEL QUESTIONS to help you in your TU IOST Entrance Examination for BSc.CSIT Admission. We have provided both NEW and OLD format because although the old one is obsolete, it is not completely useless. Refer it after finishing the NEW model.

MODEL QUESTIONS(OLD + NEW)

Model Questions(NEW FORMAT):
TU IOST Model(New Format) Questions
Model Questions(OLD FORMAT):
TU IOST Model(Old Format) Questions
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Tuesday, June 6, 2017

BSc. CSIT 7th Semester TU Board Exam Routine 2074

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Old Notices
No older notices available

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Thursday, June 1, 2017

The CSIT Song

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Couldn't resist sharing. Original song at this link
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Saturday, April 1, 2017

Some common Numerical Method Programs for 3rd Semester BSc. CSIT students

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Pro tip: If you are searching for a speciic program, hit Ctrl+F to search faster. This is quite a long list.

1. Program to find Rootof a fuction by using Bisection method
/*program to find Rootof a fuction by using Bisection method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 100
#define E 0.00001
#define f(x) x*x-4*x-10
void main()
{
 clrscr();
   int count=0;
   float x0,x1,x2,f0,f1,f2;
   printf("Enter the value of x1\n");
   scanf("%f",&x1);
   printf("Enter the value of x2\n");
   scanf("%f",&x2);
   f1=f(x1);
   f2=f(x2);
   if (f1*f2>0)
   printf("Solution does not exit");
   else
   {
 begin:
 x0=(x1+x2)/2;
 f0=f(x0);
 if((f1*f0)<0)
  x2=x0;
      else
      {
 x1=x0;
  f1=f0;
      }
      if((fabs(x2-x1)/x1)<E)
      {
 x0=(x1+x2)/2;
  printf("\nRoor=%f",x0);
  printf("\nFuctional value =%f",count);
  printf("\nIteration steps =%d",count);
      }
      else
      {
 count ++;
  goto begin;
      }
   }
   getch();
}
  

2. Program to find Rootof a fuction by using Newton Raption method
/*program to find Rootof a fuction by using Newton Raption method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 10
#define E 0.0001
#define f(x)  x*x*x-3*x-2
#define fd(x)  3*x*x-3
void main()
{
    clrscr();
    int count=0;
    float x0,x1,fx,fdx;
    printf("\nEnter the initial value of x0\n");
    scanf("%f",&x0);
begin:
    printf("\nStep %d",count+1);
    fx=f(x0);
    printf("\nvalue of fx%d=%f",count,fx);
    fdx=fd(x0);
    printf("\nvalue of fdx%d= %f",count,fdx);
    x1=x0-(fx/fdx);
    if((fabs(x1-x0)/x1)<E)
    {
        printf("\nRoot=%f",x1);
        printf("\nThe fuction value=%f",f(x1));
        printf("\nNo. of Iritation steps=%d",count);
    }
    else
    {
        x0=x1;
        count ++;
        if (count<MAX)
        {
            goto begin;
        }
        else
        {
            printf("It takes more than 10 steps So solution does not coverse");
        }
    }
    getche();
}

  

2. Program to Rootof a fuction by using Newton Raption method (Alternate)
/*program to find Rootof a fuction by using Newton Raption method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 10
#define E 0.0001
#define f(x) x*x-3*x+2
#define fd(x) 2*x-3
void main()
{
    clrscr();
    int count=0;
    float x0,x1,fx,fdx;
    printf("\nEnter the initial value of x0");
    scanf("%f",&x0);
begin:
    fx=f(x0);
    fdx=fd(x0);
    x1=x0-(fx/fdx);
    if((fabs(x1-x0)/x1)<E)
    {
        printf("\nRoot=%f",x1);
        printf("\nThe fuction value=%f",f(x1));
        printf("\nNo. of Iritation steps=%d",count);
    }
    else
    {
        x0=x1;
        count ++;
        if (count<MAX)
        {
            goto begin;
        }
        else
        {
            printf("It takes more than 10 steps So solution does not coverse");
        }
    }
    getche();
}

  

3. Program to find Root of a fuction by using Secant method
/*program to find Root of a fuction by using Secant method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 10
#define E 0.0001
#define f(x) x*x-4*x-10
void main()
{
    clrscr();
    int count=0;
    float x1,x2,x3,f1,f2;
    printf("\nEnter the two value of x1 and x2\n");
    scanf("%f%f",&x1,&x2);
begin:
    f1=f(x1);
    f2=f(x2);
    x3=x2-f2*(x2-x1)/(f2-f1);
    if((fabs(x3-x2)/x2)<E)
    {
        printf("\nRoot=%f",x3);
        printf("\nNo. of Iteration steps=%d",count);
    }
    else
    {
        x1=x2;
        x2=x3;
        f1=f2;
        f2=f(x3);
        count ++;
        if (count<MAX)
        {
            goto begin;
        }
        else
        {
            printf("It solution does not coverge");
        }

    }
    getche();
}

  

4. Program to find Rootof a fuction by using Fixed point method
/*program to find Rootof a fuction by using Fixed point method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 100
#define E 0.001
#define g(x) exp(x)/3
void main()
{
    clrscr();
    int count=0;
    float x0,x1,error;
    printf("\nEnter the initial guess value of x0\n");
    scanf("%f",&x0);
begin:
    x1=g(x0);
    error=fabs(x1-x0);
    if(error<=E)
    {
        printf("\nThe root is %f",x1);
        printf("\nThe error value %f",error);
        printf("\nThe iteration steps is %d",count);
    }
    else
    {
        x0=x1;
        count ++;
        if (count<MAX)
        {
            goto begin;
        }
        else
        {
            printf("It solution does not coverge");
        }

    }
    getch();
}

  

5. Demonstrate Lagranges Interpolation
//Lagranges interpolation
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 10
void main()
{
    clrscr();
    int i,j,n;
    float x[MAX],f[MAX],sum=0,pro,xp,fp;
    printf("Enter the number of data points:\n");
    printf("\nn= ");
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        printf("\nInput the data points for x[%d]&f[%d]",i,i);
        printf("\nx[%d]= ",i);
        scanf("%f",&x[i]);
        printf("\nf[%d]= ",i);
        scanf("%f",&f[i]);
    }
    printf("\nInput the specied value of x:");
    scanf("%f",&xp);
    for(i=0; i<=n; i++)
    {
        pro=1;
        for(j=0; j<=n; j++)
        {
            if(i!=j)
                pro=pro*(xp-x[j])/(x[i]-x[j]);
        }
        sum=sum+pro*f[i];
    }
    fp=sum;
    printf("\nThe required functional value at %f=%f",xp,fp);
    getch();
}

  

6. Demonstrate Newton Interpolation
// NEWTON INTERPOLATION
#include<stdio.h>
#include<math.h>
#include<conio.h>
#define MAX 10
void main()
{
    int i,j,n;
    float xp,fp,pro,sum,a[MAX],f[MAX],x[MAX],d[MAX][MAX];
    printf("\nInput the number of data points:");
    printf("\nn=");
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        printf("\nInput the data of x[%d]&f[%d]:",i,i);
        printf("\nx[%d]=",i);
        scanf("%f",&x[i]);
        printf("\nf[%d]=",i);
        scanf("%f",&f[i]);
    }
    for(i=1; i<=n; i++)
        d[i][1]=f[i];
    for(j=2; j<=n; j++)
        for(i=1; i<=n-j+1; i++)
            d[i][j]=(d[i+1][j-1]-d[i][j-1])/(x[i+j-1]-x[i]);
    for(j=1; j<=n; j++)
        a[j]=d[1][j];
    printf("\nInput the value of xp point:");
    printf("\nxp=");
    scanf("%f",&xp);
    sum=a[1];
    for(i=2; i<=n; i++)
    {
        pro= 1.0;
        for(j=1; j<=i-1; j++)
            pro=pro*(xp-x[j]);
        sum=sum+a[i]*pro;
    }
    fp=sum;
    printf("\n At xp=%f is %f",xp,fp);
    getch();
}
  

7. Demonstrate Least Square Method
// Least square method
#include<stdio.h>
#include<math.h>
#include<conio.h>
#define MAX 10
void main()
{
    int i,n;
    float a0,a1;
    float x[MAX],y[MAX],sumx=0.0,sumy=0.0,sumxy=0.0,sumxsq=0.0;
    printf("***Least square method***" );
    printf("\nInput the number of data points:");
    printf("\nn=");
    scanf("%d",&n);
    printf("Enter the data sets one after another:");
    for(i=0; i<n; i++)
    {
        // printf("\nInput the data of x[%d]&f[%d]:",i,i);
        printf("\nx[%d]=",i);
        scanf("%f",&x[i]);
        printf("\ny[%d]=",i);
        scanf("%f",&y[i]);
        sumx+=x[i];
        sumy+=y[i];
        sumxy+=x[i]*y[i];
        sumxsq+=x[i]*x[i];
    }
    a0=(sumy*sumxsq-sumx*sumxy)/(n*sumxsq-pow(sumx,2));
    a1=(n*sumxy-sumx*sumy)/(n*sumxsq-pow(sumx,2));
    printf("The equation of line:\n");
    printf("\n y=%fx+%f",a1,a0);
    getch();
}
  

8. Program to find Integration of given function by using Trapezoidal rule
/*program to find Integration of given funtion by using Trapezoidal rule*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define f(x) x*x*x+1
void main()
{
    clrscr();
    float a,b,h,It;
    printf("\nEnter initial value of X:");
    scanf("%f",&a);
    printf("\nEnter  Final  value of X:");
    scanf("%f",&b);
    h=(b-a)/2;
    It =h*(f(a)+f(b));
    printf("\nThe integration value of function :%f",It);
    getch();
}

  

9. Program to find Integration of given tabulated data using Composite Trapezoidal rule*
/*program to find Integration of given tabulated data using  Composit Trapezoidal rule*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define max 15
void main()
{
    clrscr();
    int n,n1,n2,i;
    float a,b,h,sum,ict,x[max],y[max];
    printf("\nEnter the number of data points:");
    printf("\nn=");
    scanf("%d",&n);
    printf("\nInput the values set by set");
    for(i=1; i<=n; i++)
    {
        printf("\nX%i=",i);
        scanf("%f",&x[i]);
        printf("\nF(x%i)=",i);
        scanf("%f",&y[i]);
    }
    printf("\nInput the initial value.");
    printf("\na=");
    scanf("%f",&a);
    printf("\nInput the final value.");
    printf("\nb=");
    scanf("%f",&b);
    h=x[2]-x[1];
    n1= (int)(fabs(a-x[1])/h+1.5);
    n2= (int)(fabs(b-x[1])/h+1.5) ;
    sum=0.0;
    for(i=n1; i<=n2; i++)
        sum =sum+y[i]+y[i+1];
    ict= sum*h/2.0;
    printf("\nIntegral from %f to %f is %f\n",a,b,ict);
    getche();
}
  

10. Program to find Integration of given funtion by using Composite Trapezoidal rule
/*program to find Integration of given funtion by using composit Trapezoidal rule*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
    clrscr();
    int i,n;
    float a,b,h,sum,Ict;
    float F(float x);
    printf("Enter the initial value of a:");
    scanf("%f",&a);
    printf("\nEnter the Final  value of b: ");
    scanf("%f",&b);
    printf("\nEnter the segment width:");
    scanf("%f",&h);
    n=(b-a)/h;
    sum =(F(a)+F(b))/2.0;
    for(i=1; i<n-1; i++)
    {
        sum= sum +F(a+i*h);
    }
    Ict=sum*h;
    printf("\nThe integration between %f and %f at h=%f is %f",a,b,h,Ict);
    getch();
}
float F(float x)
{
    float f;
    f=1-exp(-x/2.0) ;
    return (f);
}

  

11. Program to find Integration of given funtion by using Simson's 1/3 rule
/*program to find Integration of given funtion by using Simson's 1/3 rule*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define F(x) 1-exp(-(x)/2.0)
void main()
{
 clrscr();
   int n,m,i;
   float a,b,h,sum,Ics,x,f1,f2,f3;
   printf("Enter initial value of X:");
   printf("\na=");scanf("%f",&a);
   printf("\nEnter  Final  value of X:");
   printf("\nb=");scanf("%f",&b);
   printf("\nEnter number of segments (Even number):");
   printf("\nN=");scanf("%d",&n);
   h=(b-a)/n;
   m=n/2;
   sum =0.0;
   x=a;
   f1=F(x);
   for(i=1;i<m;i++)
   {
    f2=F(x+h);
      f3=F(x+2*h);
      sum=sum+f1+4*f2+f3;
      f1=f3;
      x=x+2*h;
   }
   Ics=sum *h/3.0;
   printf("\nIntegral from %f to %f\n",a,b);
   printf("When h=%f is %f \n",h,Ics);
   getch();
}
  

12. Program to find Integration of given funtion by using Simson's 1/8 rule
/*program to find Integration of given funtion by using Simson's 1/8 rule*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define F(x) (x*x*x+1)
void main()
{
    clrscr();
    float a,b,h,Is;
    printf("\nEnter initial value of X:");
    scanf("%f",&a);
    printf("\nEnter  Final  value of X:");
    scanf("%f",&b);
    h=(b-a)/3;
    Is=3*h*(F(a)+3*F(a+h)+3*F(a+2*h)+F(b))/8;
    printf("\nIntegral from %f to %f\n",a,b);
    printf("Integration is %f \n",Is);
    getch();
}

  

13. Program to solve Integration equation by using Gauss Integration Method for n=2 interpolating point
/*program to solve Integration equation by using gauss integration method for
n=2 interpolating point*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define f(x) exp(x)
void main()
{
    clrscr();
    int n=2,i;
    float a,b,w[2],z[2],x[2],Ig =0.0;
    printf("\nInput initial value of x:");
    printf("\na=");
    scanf("%f",&a);
    printf("\nInput final value of x:");
    printf("\nb=");
    scanf("%f",&b);
    w[0]=1.0;
    w[1]=1.0;
    z[0]=-0.5773502;
    z[1]=0.5773502;
    for (i=0; i<n; i++)
        x[i]=((b-a)/2)* z[i]+((b+a)/2);
    for(i=0; i<n; i++)
        Ig +=((b-a)/2)*(w[i]*f(z[i]));
    printf("\nIntegral from %f to %f is %f\n",a,b,Ig);
    getche();
}

  

14. Program to solve Integration equation by using gauss integration method for n=3 interpolating point
/*program to solve Integration equation by using gauss integration method for
n=3 interpolating point*/
#include&lt;stdio.h&gt;
#include&lt;conio.h&gt;
#include&lt;math.h&gt;
#define f(x) exp(x)
void main()
{
    clrscr();
    int n=3,i;
    float a,b,w[3],z[3],x[3],Ig =0.0000;
    printf("\nInput initial value of x:");
    printf("\na=");
    scanf("%f",&amp;a);
    printf("\nInput final value of x:");
    printf("\nb=");
    scanf("%f",&amp;b);
    w[0]=0.55556;
    w[1]=0.88889;
    w[2]=0.55556;
    z[0]=-0.77460;
    z[1]=0.00;
    z[2]=0.77460;
    for (i=0; i&lt;n; i++)
        x[i]=((b-a)/2)* z[i]+((b+a)/2);
    for(i=0; i&lt;n; i++)
        Ig +=((b-a)/2)*(w[i]*f(z[i]));
    printf("\nIntegral from %f to %f is %f\n",a,b,Ig);
    getche();
}
  

15. Demonstrate Euler Method
//Eluer method
#include <stdio.h>
#include <conio.h>
#include <math.h>
#define f(x,y) 2*y/x
void main()
{
    clrscr();
    int i,n;
    float x,y,h,xp,dy;
    printf("Input initial value of x and y");
    scanf ("%f%f",&x,&y);
    printf("Input X-value at which Y is required :\n");
    scanf("%f",&xp);
    printf("Input  step size:\n");
    scanf("%f",&h);
    n=((xp-x)/h+0.5);
    for (i=1; i<=n; i++)
    {
        dy=h*f(x,y);
        x=x+h;
        y=y+dy;
        printf("x=%f and y=%f\n",x,y);
    }
    printf("\n Value of y at x =%f is %f",x,y);
    getche();
}

  

16. Demonstrate Heuns Method
// Heuns method
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define max 15
void main()
{
    clrscr();
    int n,i;
    float x,y,xp,h,m1,m2;
    float func(float,float);
    printf("\n***Solution of Heuns method***");
    printf("\nEnter the initial value of x \n");
    scanf("%f",&x);
    printf("\nEnter the initial value of y \n");
    scanf("%f",&y);
    printf("\nInput x at which y is required");
    scanf("%f",&xp);
    printf("\nInput step size");
    scanf("%f",&h);
    n=(int)((xp-x)/h+0.5);
    for(i=1; i<=n; i++)
    {
        m1=func(x,y);
        m2=func(x+h,y+m1*h);
        x=x+h;
        y=y+0.5*h*(m1+m2);
        printf("%5d %10.6f %10.6f\n",i,x,y);
    }
    printf("\nThe value of y at x=%f is %f\n",x,y);
    getch();
}
float func(float x,float y)
{
    float f;
    f=2.0*y/x;
    return(f);
}

  

17. Demonstrate Runge-Kutta Method
#include <stdio.h>
#include <conio.h>
#include <math.h>
//#define f(x,y) (2.0*y/x)
void main()
{
  clrscr();
  int n,i;
  float x,y,xp,h;
  float m1,m2,m3,m4;
  float func(float , float );
  printf("Input initial value of x and y");
  scanf ("%f%f",&x,&y);
  printf("Input X-value at which Y is required :\n");
  scanf("%f",&xp);
  printf("Input  step size:\n");
  scanf("%f",&h);
  n=(int)((xp-x)/h+0.5);
  for (i=1;i<=n;i++)
  {
    m1=func(x,y);
      m2=func(x+0.5*h,y+0.5+m1*h);
      m3=func(x+0.5*h,y+0.5+m2*h);
      m4=func(x+h,y+m3*h);
      x=x+h;
      y=y+(m1+2.0*m2+2.0*m3+m4)*h/6.0;
      printf("x=%f and y=%f\n",x,y);
  }
  printf("\n Value of y at x=%f is %f",x,y);
    getche();
}
float func(float x, float y)
  {
    float f;
      f=x*x+y*y;
      return(f);
  }
  

18. Runge-Kutta (Alternate)
#include <stdio.h>
#include <conio.h>
#include <math.h>
#define f(x,y) 2.0*y/x
void main()
{
    clrscr();
    int i,n;
    float x,y,xp,h;
    float m1,m2,m3,m4;
    printf("Input initial value of x and y");
    scanf ("%f%f",&x,&y);
    printf("Input X-value at which Y is required :\n");
    scanf("%f",&xp);
    printf("Input  step size:\n");
    scanf("%f",&h);
    n=(int)((xp-x)/h+0.5);
    for (i=1; i<=n; i++)
    {
        m1=f(x,y);
        m2=f(x+0.5*h,y+(m1*h*0.5));
        m3=f(x+0.5*h,y+(m2*h*0.5));
        m4=f(x+h,y+m3*h);
        x=x+h;
        y=y+((m1+2.0*m2+2.0*m3+m4)*h)/6.0;
        printf("x=%f and y=%f\n",x,y);
    }
    printf("\n Value of y at xp =%f is %f",x,y);
    getche();
}

  

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Program to find Root of a fuction by using Newton Raphson method

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Program to find Root of a fuction by using Newton Raphson method
/*program to find Rootof a fuction by using Newton Raphson method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 10
#define E 0.0001
#define f(x)  x*x*x-3*x-2
#define fd(x)  3*x*x-3
void main()
{
    clrscr();
    int count=0;
    float x0,x1,fx,fdx;
    printf("\nEnter the initial value of x0\n");
    scanf("%f",&x0);
begin:
    printf("\nStep %d",count+1);
    fx=f(x0);
    printf("\nvalue of fx%d=%f",count,fx);
    fdx=fd(x0);
    printf("\nvalue of fdx%d= %f",count,fdx);
    x1=x0-(fx/fdx);
    if((fabs(x1-x0)/x1)<E)
    {
        printf("\nRoot=%f",x1);
        printf("\nThe fuction value=%f",f(x1));
        printf("\nNo. of Iritation steps=%d",count);
    }
    else
    {
        x0=x1;
        count ++;
        if (count<MAX)
        {
            goto begin;
        }
        else
        {
            printf("It takes more than 10 steps So solution does not coverse");
        }
    }
    getche();
}

  

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Program to find Root of a fuction by using Bisection method

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Program to find Rootof a fuction by using Bisection method
/*program to find Rootof a fuction by using Bisection method*/
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define MAX 100
#define E 0.00001
#define f(x) x*x-4*x-10
void main()
{
  clrscr();
   int count=0;
   float x0,x1,x2,f0,f1,f2;
   printf("Enter the value of x1\n");
   scanf("%f",&x1);
   printf("Enter the value of x2\n");
   scanf("%f",&x2);
   f1=f(x1);
   f2=f(x2);
   if (f1*f2>0)
   printf("Solution does not exit");
   else
   {
  begin:
  x0=(x1+x2)/2;
  f0=f(x0);
  if((f1*f0)<0)
    x2=x0;
      else
      {
  x1=x0;
   f1=f0;
      }
      if((fabs(x2-x1)/x1)<E)
      {
  x0=(x1+x2)/2;
   printf("\nRoor=%f",x0);
   printf("\nFuctional value =%f",count);
   printf("\nIteration steps =%d",count);
      }
      else
      {
  count ++;
   goto begin;
      }
   }
   getch();
}
  

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