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Differential Equation

87613 One root of the equation $x^{3} - 4 x + 1 = 0$ is between 1 and 2 . The value of this root using Newton- Raphson method will be

1 1.775

2 1.850

3 1.875

4 1.950

Explanation:

Exp:(c) we have,
$f (x) = x^{3} - 4 x + 1$
$∴ f^{'} (x) = 3 x^{2} - 4$
Let, $x_{0} = 2$
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$x_{1} = 2 - \frac{1}{8} = \frac{16 - 1}{8} = \frac{15}{8} = 1.875$

CG PET- 2017

Differential Equation

87614 Dividing the interval $(1, 2)$ into four equal parts and using Simpson's rule, the value of $\int_{1}^{2} \frac{d x}{x}$ will be

1 0.6932

2 0.6753

3 0.6692

4 0.7132

Explanation:

Exp:(a)
We have $\int_{1}^{2} \frac{d x}{x}$
Here, $a = 1, b = 2, n = 4$
$∴ nh = b - a$
$\Rightarrow 4 h = 2 - 1 = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{x}$ are given below
| $x$ | 1 | $\frac{5}{4}$ | $\frac{6}{4}$ | $\frac{7}{4}$ | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By Simpson's rule
$\int_{1}^{2} \frac{dx}{x} = \frac{h}{3} [(y_{0} + y_{4}) + 4 (y_{1} + y_{3}) + 2 (y_{2})]$
$= \frac{1}{12} [(1 + \frac{1}{2}) + 4 (\frac{4}{5} + \frac{4}{7}) + 2 (\frac{4}{6})]$
$= \frac{1}{12} [\frac{3}{2} + 16 (\frac{12}{35}) + \frac{4}{3}] = \frac{1}{8} + \frac{16}{35} + \frac{1}{9}$
$= 0.125 + 0.4571 + 0.1111 = 0.6932$

CG PET- 2017

Differential Equation

87615 Taking four subintervals, the values of $\int_{0}^{1} \frac{d x}{1 + x}$ by using trapezoidal rule will be

1 0.6870

2 0.6677

3 0.6970

4 0.5970

Explanation:

(C) : We have, $\int_{0}^{1} \frac{dx}{1 + x}$
Here, $a = 0, b = 1, n = 4$
Now, $nh = b - a$
$4 h = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{1 + x}$ are given below
| $x$ | 0 | $\frac{1}{4}$ | $\frac{2}{4}$ | $\frac{3}{4}$ | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By trapezoidal rule
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{h}{2} [(y_{0} + y_{4}) + 2 (y_{1} + y_{2} + y_{3})]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{8} [(1 + \frac{1}{2}) + 2 (\frac{4}{5} + \frac{4}{6} + \frac{4}{7})]$
$= \frac{1}{8} [\frac{3}{2} + 8 (\frac{42 + 35 + 30}{210})]$
$= \frac{3}{16} + \frac{107}{210}$
$= 0.1875 + 0.5095 = 0.6970$

CG PET- 2017

Differential Equation

87616 One root of the equation, $x^{3} - 3 x - 5 = 0$ lies between 2 and 2.5. Using Newton-Raphson method, the value of that root will be

1 2.25

2 2.33

3 2.35

4 2.45

Explanation:

(B) : $Let f (x) = x^{3} - 3 x - 5 = 0$
Then $f^{'} (x) = 3 x^{2} - 3$
Using Newton Raphson formula
$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})} = x_{n} - \frac{x_{n}^{3} - 3 x_{n} - 5}{3 x_{n}^{2} - 3}$
$x_{n + 1} = \frac{2 x_{n}^{3} + 5}{3 x_{n}^{2} - 3}$
Given, one root is between 2 and 2.5 so taking $x_{0} = 2$ , $n = 0$ in equation (i)
$x_{1} = \frac{2 x_{0}^{3} + 5}{3 x_{0}^{2} - 3} = \frac{2 (2)^{3} + 5}{3 (2)^{2} - 3} = \frac{21}{9} = 2.33$

CG PET- 2018

Differential Equation

87613 One root of the equation $x^{3} - 4 x + 1 = 0$ is between 1 and 2 . The value of this root using Newton- Raphson method will be

1 1.775

2 1.850

3 1.875

4 1.950

Explanation:

Exp:(c) we have,
$f (x) = x^{3} - 4 x + 1$
$∴ f^{'} (x) = 3 x^{2} - 4$
Let, $x_{0} = 2$
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$x_{1} = 2 - \frac{1}{8} = \frac{16 - 1}{8} = \frac{15}{8} = 1.875$

CG PET- 2017

Differential Equation

87614 Dividing the interval $(1, 2)$ into four equal parts and using Simpson's rule, the value of $\int_{1}^{2} \frac{d x}{x}$ will be

1 0.6932

2 0.6753

3 0.6692

4 0.7132

Explanation:

Exp:(a)
We have $\int_{1}^{2} \frac{d x}{x}$
Here, $a = 1, b = 2, n = 4$
$∴ nh = b - a$
$\Rightarrow 4 h = 2 - 1 = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{x}$ are given below
| $x$ | 1 | $\frac{5}{4}$ | $\frac{6}{4}$ | $\frac{7}{4}$ | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By Simpson's rule
$\int_{1}^{2} \frac{dx}{x} = \frac{h}{3} [(y_{0} + y_{4}) + 4 (y_{1} + y_{3}) + 2 (y_{2})]$
$= \frac{1}{12} [(1 + \frac{1}{2}) + 4 (\frac{4}{5} + \frac{4}{7}) + 2 (\frac{4}{6})]$
$= \frac{1}{12} [\frac{3}{2} + 16 (\frac{12}{35}) + \frac{4}{3}] = \frac{1}{8} + \frac{16}{35} + \frac{1}{9}$
$= 0.125 + 0.4571 + 0.1111 = 0.6932$

CG PET- 2017

Differential Equation

87615 Taking four subintervals, the values of $\int_{0}^{1} \frac{d x}{1 + x}$ by using trapezoidal rule will be

1 0.6870

2 0.6677

3 0.6970

4 0.5970

Explanation:

(C) : We have, $\int_{0}^{1} \frac{dx}{1 + x}$
Here, $a = 0, b = 1, n = 4$
Now, $nh = b - a$
$4 h = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{1 + x}$ are given below
| $x$ | 0 | $\frac{1}{4}$ | $\frac{2}{4}$ | $\frac{3}{4}$ | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By trapezoidal rule
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{h}{2} [(y_{0} + y_{4}) + 2 (y_{1} + y_{2} + y_{3})]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{8} [(1 + \frac{1}{2}) + 2 (\frac{4}{5} + \frac{4}{6} + \frac{4}{7})]$
$= \frac{1}{8} [\frac{3}{2} + 8 (\frac{42 + 35 + 30}{210})]$
$= \frac{3}{16} + \frac{107}{210}$
$= 0.1875 + 0.5095 = 0.6970$

CG PET- 2017

Differential Equation

87616 One root of the equation, $x^{3} - 3 x - 5 = 0$ lies between 2 and 2.5. Using Newton-Raphson method, the value of that root will be

1 2.25

2 2.33

3 2.35

4 2.45

Explanation:

(B) : $Let f (x) = x^{3} - 3 x - 5 = 0$
Then $f^{'} (x) = 3 x^{2} - 3$
Using Newton Raphson formula
$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})} = x_{n} - \frac{x_{n}^{3} - 3 x_{n} - 5}{3 x_{n}^{2} - 3}$
$x_{n + 1} = \frac{2 x_{n}^{3} + 5}{3 x_{n}^{2} - 3}$
Given, one root is between 2 and 2.5 so taking $x_{0} = 2$ , $n = 0$ in equation (i)
$x_{1} = \frac{2 x_{0}^{3} + 5}{3 x_{0}^{2} - 3} = \frac{2 (2)^{3} + 5}{3 (2)^{2} - 3} = \frac{21}{9} = 2.33$

CG PET- 2018

Differential Equation

87613 One root of the equation $x^{3} - 4 x + 1 = 0$ is between 1 and 2 . The value of this root using Newton- Raphson method will be

1 1.775

2 1.850

3 1.875

4 1.950

Explanation:

Exp:(c) we have,
$f (x) = x^{3} - 4 x + 1$
$∴ f^{'} (x) = 3 x^{2} - 4$
Let, $x_{0} = 2$
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$x_{1} = 2 - \frac{1}{8} = \frac{16 - 1}{8} = \frac{15}{8} = 1.875$

CG PET- 2017

Differential Equation

87614 Dividing the interval $(1, 2)$ into four equal parts and using Simpson's rule, the value of $\int_{1}^{2} \frac{d x}{x}$ will be

1 0.6932

2 0.6753

3 0.6692

4 0.7132

Explanation:

Exp:(a)
We have $\int_{1}^{2} \frac{d x}{x}$
Here, $a = 1, b = 2, n = 4$
$∴ nh = b - a$
$\Rightarrow 4 h = 2 - 1 = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{x}$ are given below
| $x$ | 1 | $\frac{5}{4}$ | $\frac{6}{4}$ | $\frac{7}{4}$ | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By Simpson's rule
$\int_{1}^{2} \frac{dx}{x} = \frac{h}{3} [(y_{0} + y_{4}) + 4 (y_{1} + y_{3}) + 2 (y_{2})]$
$= \frac{1}{12} [(1 + \frac{1}{2}) + 4 (\frac{4}{5} + \frac{4}{7}) + 2 (\frac{4}{6})]$
$= \frac{1}{12} [\frac{3}{2} + 16 (\frac{12}{35}) + \frac{4}{3}] = \frac{1}{8} + \frac{16}{35} + \frac{1}{9}$
$= 0.125 + 0.4571 + 0.1111 = 0.6932$

CG PET- 2017

Differential Equation

87615 Taking four subintervals, the values of $\int_{0}^{1} \frac{d x}{1 + x}$ by using trapezoidal rule will be

1 0.6870

2 0.6677

3 0.6970

4 0.5970

Explanation:

(C) : We have, $\int_{0}^{1} \frac{dx}{1 + x}$
Here, $a = 0, b = 1, n = 4$
Now, $nh = b - a$
$4 h = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{1 + x}$ are given below
| $x$ | 0 | $\frac{1}{4}$ | $\frac{2}{4}$ | $\frac{3}{4}$ | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By trapezoidal rule
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{h}{2} [(y_{0} + y_{4}) + 2 (y_{1} + y_{2} + y_{3})]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{8} [(1 + \frac{1}{2}) + 2 (\frac{4}{5} + \frac{4}{6} + \frac{4}{7})]$
$= \frac{1}{8} [\frac{3}{2} + 8 (\frac{42 + 35 + 30}{210})]$
$= \frac{3}{16} + \frac{107}{210}$
$= 0.1875 + 0.5095 = 0.6970$

CG PET- 2017

Differential Equation

87616 One root of the equation, $x^{3} - 3 x - 5 = 0$ lies between 2 and 2.5. Using Newton-Raphson method, the value of that root will be

1 2.25

2 2.33

3 2.35

4 2.45

Explanation:

(B) : $Let f (x) = x^{3} - 3 x - 5 = 0$
Then $f^{'} (x) = 3 x^{2} - 3$
Using Newton Raphson formula
$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})} = x_{n} - \frac{x_{n}^{3} - 3 x_{n} - 5}{3 x_{n}^{2} - 3}$
$x_{n + 1} = \frac{2 x_{n}^{3} + 5}{3 x_{n}^{2} - 3}$
Given, one root is between 2 and 2.5 so taking $x_{0} = 2$ , $n = 0$ in equation (i)
$x_{1} = \frac{2 x_{0}^{3} + 5}{3 x_{0}^{2} - 3} = \frac{2 (2)^{3} + 5}{3 (2)^{2} - 3} = \frac{21}{9} = 2.33$

CG PET- 2018

Differential Equation

87613 One root of the equation $x^{3} - 4 x + 1 = 0$ is between 1 and 2 . The value of this root using Newton- Raphson method will be

1 1.775

2 1.850

3 1.875

4 1.950

Explanation:

Exp:(c) we have,
$f (x) = x^{3} - 4 x + 1$
$∴ f^{'} (x) = 3 x^{2} - 4$
Let, $x_{0} = 2$
$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})}$
$x_{1} = 2 - \frac{1}{8} = \frac{16 - 1}{8} = \frac{15}{8} = 1.875$

CG PET- 2017

Differential Equation

87614 Dividing the interval $(1, 2)$ into four equal parts and using Simpson's rule, the value of $\int_{1}^{2} \frac{d x}{x}$ will be

1 0.6932

2 0.6753

3 0.6692

4 0.7132

Explanation:

Exp:(a)
We have $\int_{1}^{2} \frac{d x}{x}$
Here, $a = 1, b = 2, n = 4$
$∴ nh = b - a$
$\Rightarrow 4 h = 2 - 1 = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{x}$ are given below
| $x$ | 1 | $\frac{5}{4}$ | $\frac{6}{4}$ | $\frac{7}{4}$ | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By Simpson's rule
$\int_{1}^{2} \frac{dx}{x} = \frac{h}{3} [(y_{0} + y_{4}) + 4 (y_{1} + y_{3}) + 2 (y_{2})]$
$= \frac{1}{12} [(1 + \frac{1}{2}) + 4 (\frac{4}{5} + \frac{4}{7}) + 2 (\frac{4}{6})]$
$= \frac{1}{12} [\frac{3}{2} + 16 (\frac{12}{35}) + \frac{4}{3}] = \frac{1}{8} + \frac{16}{35} + \frac{1}{9}$
$= 0.125 + 0.4571 + 0.1111 = 0.6932$

CG PET- 2017

Differential Equation

87615 Taking four subintervals, the values of $\int_{0}^{1} \frac{d x}{1 + x}$ by using trapezoidal rule will be

1 0.6870

2 0.6677

3 0.6970

4 0.5970

Explanation:

(C) : We have, $\int_{0}^{1} \frac{dx}{1 + x}$
Here, $a = 0, b = 1, n = 4$
Now, $nh = b - a$
$4 h = 1$
$h = \frac{1}{4}$
The value of $f (x) = \frac{1}{1 + x}$ are given below
| $x$ | 0 | $\frac{1}{4}$ | $\frac{2}{4}$ | $\frac{3}{4}$ | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $f (x)$ | $y_{0} = 1$ | $y_{1} = \frac{4}{5}$ | $y_{2} = \frac{4}{6}$ | $y_{3} = \frac{4}{7}$ | $y_{4} = \frac{1}{2}$ |
By trapezoidal rule
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{h}{2} [(y_{0} + y_{4}) + 2 (y_{1} + y_{2} + y_{3})]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{8} [(1 + \frac{1}{2}) + 2 (\frac{4}{5} + \frac{4}{6} + \frac{4}{7})]$
$= \frac{1}{8} [\frac{3}{2} + 8 (\frac{42 + 35 + 30}{210})]$
$= \frac{3}{16} + \frac{107}{210}$
$= 0.1875 + 0.5095 = 0.6970$

CG PET- 2017

Differential Equation

87616 One root of the equation, $x^{3} - 3 x - 5 = 0$ lies between 2 and 2.5. Using Newton-Raphson method, the value of that root will be

1 2.25

2 2.33

3 2.35

4 2.45

Explanation:

(B) : $Let f (x) = x^{3} - 3 x - 5 = 0$
Then $f^{'} (x) = 3 x^{2} - 3$
Using Newton Raphson formula
$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})} = x_{n} - \frac{x_{n}^{3} - 3 x_{n} - 5}{3 x_{n}^{2} - 3}$
$x_{n + 1} = \frac{2 x_{n}^{3} + 5}{3 x_{n}^{2} - 3}$
Given, one root is between 2 and 2.5 so taking $x_{0} = 2$ , $n = 0$ in equation (i)
$x_{1} = \frac{2 x_{0}^{3} + 5}{3 x_{0}^{2} - 3} = \frac{2 (2)^{3} + 5}{3 (2)^{2} - 3} = \frac{21}{9} = 2.33$

CG PET- 2018