Q1) Find a real root of using bisection method correct to five decimal places.

A1)

Find a real root of using bisection method correct to five decimal places.

Let then by hit and trial we have

Thus .So the root of the given equation should lies between 1 and 2.

Now,

I.e. positive so the root of the given equation must lies between

Now,

I.e. negative so the root of the given equation lies between

Now,

i.e. positive so the root of the given equation lies between

Now,

i.e. negative so that the root of the given equation lies between

Now,

i.e. positive so that the root of the given equation lies between

Now,

i.e. positive so that the root of the given equation lies between

Now,

i.e. negative so that the root of the given equation lies between

Now,

i.e. negative so that the root of the given equation lies between

Hence the approximate root of the given equation is 1.32421

Q2) Find the root of the equation   between 2 and 3, using bisection method correct to two decimal places.

A2)

Let

Where

Thus .So the root of the given equation should lie between 2 and 3.

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. negative so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Now,

i.e. positive so the root of the given equation must lie between

Hence the root of the given equation correct to two decimal places is 2.1269

Q3) Find a real root of the equation near, correct to three decimal places by the Regula Falsi method.

A3)

Let

Now,

And also

Hence the root of the equation lies between and and so,

By Regula Falsi Method

Now,

So the root of the equation lies between 1 and 0.5 and so

By Regula Falsi Method

Now,

So the root of the equation lies between 1 and 0.63637 and so

By Regula Falsi Method

Now,

So the root of the equation lies between 1 and 0.67112 and so

By Regula Falsi Method

Now,

So the root of the equation lies between 1 and 0.63636 and so

By Regula Falsi Method

Now,

So the root of the equation lies between 1 and 0.68168 and so

By Regula Falsi Method

Now,

Hence the approximate root of the given equation near to 1 is 0.68217

Example 2 Find the real root of the equation

By the method of false position correct to four decimal places

Let   

By hit and trail method

0.23136 > 0

So, the root of the equation lies between 2 and 3 and also

By Regula Falsi Method

Now,

So, root of the equation   lies between 2.72101 and 3 and also

By Regula Falsi Method

Now,

So, root of the equation   lies between 2.74020 and 3 and also

By Regula Falsi Method

Now,

So, root of the equation   lies between 2.74063 and 3 and also

By Regula Falsi Method

Hence the root of the given equation correct to four decimal places is 2.7406

Q4) Find the real root of the equation

By the method of false position correct to four decimal places

A4)

Let   

By hit and trail method

0.23136 > 0

So, the root of the equation lies between 2 and 3 and also

By Regula   Falsi Method

Now,

So, root of the equation   lies between 2.72101 and 3 and also

By Regula   Falsi Method

Now,

So, root of the equation   lies between 2.74020 and 3 and also

By Regula   Falsi Method

Now,

So, root of the equation   lies between 2.74063 and 3 and also

By Regula   Falsi Method

Hence the root of the given equation correct to four decimal places is 2.7406

Q5) Apply Regula Falsi Method to solve the equation

A5)

Let 

By hit and trail

And

So, the root of the equation lies between and also

By Regula Falsi Method

Now,

So, root of the equation   lies between 0.60709 and 0.61 and also

By Regula Falsi Method

Now,

So, root of the equation   lies between 0.60710 and 0.61 and also

By Regula Falsi Method

Hence the root of the given equation correct to five decimal places is 0.60710.

Q6) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places:

A6)

Given

By Newton Raphson Method

= 

=

The initial approximation is in radian.

For n =0, the first approximation

For n =1, the second approximation

For n =2, the third approximation

For n =3, the fourth approximation

Hence the root of the given equation correct to five decimal places is 2.79838.

Q7) Using Newton-Raphson method, find a root of the following equation correct to 3 decimal places: near to 4.5

A7)

Let

The initial approximation

By Newton Raphson Method

For n =0, the first approximation

For n =1, the second approximation

For n =2, the third approximation

For n =3, the fourth approximation

Hence the root of the equation correct to three decimal places is 4.5579

Q8) Using Newton-Raphson method, find a root of the following equation correct to 4 decimal places:

A8)

Let

By Newton Raphson Method

Let the initial approximation be

For n=0, the first approximation

For n=1, the second approximation

For n=2, the third approximation

Since therefore the root of the given equation correct to four decimal places is -2.9537

Q9) Fit a straight line to the following data regarding x as the independent variables:

A9)

The   equation of straight line is

And the normal equations

We construct the data table:

Here    (no. Of steps)

Substituting the values from table in normal equations:

16.9=5a+10b

47.1=10a+30b

On solving we getand

Therefore, the required equation of the straight line is .

Q10) Predict y at x = 3.75 by fitting a power curve  to the given data
 

A10)

Given equation is 

Taking log on both sides,

Let    

Therefore, its normal equation is

We construct the data table:

Substituting the   values from the   table in the above equations

On solving we get

Hence the required equation is

So,