Miscellaneous Application of Differential Equation
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Differential Equation

87609 By Newton-Raphson method, the positive root of the equation \(x^{4}-x-10=0\) is

1 1.871
2 1.868
3 1.856
4 None of
CG PET- 2015
Differential Equation

87610 By trapezoidal rule, the approximate value of the integral \(\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\) is

1 1.3128
2 1.4108
3 1.4218
4 None of these
CG PET- 2015
Differential Equation

87618 Using trapezoidal rule and taking \(n=4\), the approximate value of integral \(\int_{1}^{9} x^{2} d x\) is
\(2\left[\frac{1}{2}\left(1+9^{2}\right)+\alpha^{2}+\beta^{2}+7^{2}\right]\) then

1 \(\alpha=1, \beta=3\)
2 \(\alpha=2, \beta=4\)
3 \(\alpha=3, \beta=5\)
4 \(\alpha=4, \beta=6\)
CG PET- 2018
Differential Equation

87611 The positive root of equation \(x^{3}-2 x-5=0\) lies in the interval

1 \((0,1)\)
2 \((1,2)\)
3 \((2,3)\)
4 \((3,4)\)
CG PET- 2016
Differential Equation

87612 The value of \(\int_{0}^{6} \frac{d x}{1+x^{2}}\) by choosing six subintervals and by using Simpson's Rule will be

1 1.3562
2 1.3662
3 1.3456
4 1.2662
CG PET- 2016
NEET DIGITAL LIBRARY
Differential Equation

87609 By Newton-Raphson method, the positive root of the equation \(x^{4}-x-10=0\) is

1 1.871
2 1.868
3 1.856
4 None of
CG PET- 2015
Differential Equation

87610 By trapezoidal rule, the approximate value of the integral \(\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\) is

1 1.3128
2 1.4108
3 1.4218
4 None of these
CG PET- 2015
Differential Equation

87618 Using trapezoidal rule and taking \(n=4\), the approximate value of integral \(\int_{1}^{9} x^{2} d x\) is
\(2\left[\frac{1}{2}\left(1+9^{2}\right)+\alpha^{2}+\beta^{2}+7^{2}\right]\) then

1 \(\alpha=1, \beta=3\)
2 \(\alpha=2, \beta=4\)
3 \(\alpha=3, \beta=5\)
4 \(\alpha=4, \beta=6\)
CG PET- 2018
Differential Equation

87611 The positive root of equation \(x^{3}-2 x-5=0\) lies in the interval

1 \((0,1)\)
2 \((1,2)\)
3 \((2,3)\)
4 \((3,4)\)
CG PET- 2016
Differential Equation

87612 The value of \(\int_{0}^{6} \frac{d x}{1+x^{2}}\) by choosing six subintervals and by using Simpson's Rule will be

1 1.3562
2 1.3662
3 1.3456
4 1.2662
CG PET- 2016
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Differential Equation

87609 By Newton-Raphson method, the positive root of the equation \(x^{4}-x-10=0\) is

1 1.871
2 1.868
3 1.856
4 None of
CG PET- 2015
Differential Equation

87610 By trapezoidal rule, the approximate value of the integral \(\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\) is

1 1.3128
2 1.4108
3 1.4218
4 None of these
CG PET- 2015
Differential Equation

87618 Using trapezoidal rule and taking \(n=4\), the approximate value of integral \(\int_{1}^{9} x^{2} d x\) is
\(2\left[\frac{1}{2}\left(1+9^{2}\right)+\alpha^{2}+\beta^{2}+7^{2}\right]\) then

1 \(\alpha=1, \beta=3\)
2 \(\alpha=2, \beta=4\)
3 \(\alpha=3, \beta=5\)
4 \(\alpha=4, \beta=6\)
CG PET- 2018
Differential Equation

87611 The positive root of equation \(x^{3}-2 x-5=0\) lies in the interval

1 \((0,1)\)
2 \((1,2)\)
3 \((2,3)\)
4 \((3,4)\)
CG PET- 2016
Differential Equation

87612 The value of \(\int_{0}^{6} \frac{d x}{1+x^{2}}\) by choosing six subintervals and by using Simpson's Rule will be

1 1.3562
2 1.3662
3 1.3456
4 1.2662
CG PET- 2016
For More Updates join with Us
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Differential Equation

87609 By Newton-Raphson method, the positive root of the equation \(x^{4}-x-10=0\) is

1 1.871
2 1.868
3 1.856
4 None of
CG PET- 2015
Differential Equation

87610 By trapezoidal rule, the approximate value of the integral \(\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\) is

1 1.3128
2 1.4108
3 1.4218
4 None of these
CG PET- 2015
Differential Equation

87618 Using trapezoidal rule and taking \(n=4\), the approximate value of integral \(\int_{1}^{9} x^{2} d x\) is
\(2\left[\frac{1}{2}\left(1+9^{2}\right)+\alpha^{2}+\beta^{2}+7^{2}\right]\) then

1 \(\alpha=1, \beta=3\)
2 \(\alpha=2, \beta=4\)
3 \(\alpha=3, \beta=5\)
4 \(\alpha=4, \beta=6\)
CG PET- 2018
Differential Equation

87611 The positive root of equation \(x^{3}-2 x-5=0\) lies in the interval

1 \((0,1)\)
2 \((1,2)\)
3 \((2,3)\)
4 \((3,4)\)
CG PET- 2016
Differential Equation

87612 The value of \(\int_{0}^{6} \frac{d x}{1+x^{2}}\) by choosing six subintervals and by using Simpson's Rule will be

1 1.3562
2 1.3662
3 1.3456
4 1.2662
CG PET- 2016
NEET DIGITAL LIBRARY
Differential Equation

87609 By Newton-Raphson method, the positive root of the equation \(x^{4}-x-10=0\) is

1 1.871
2 1.868
3 1.856
4 None of
CG PET- 2015
Differential Equation

87610 By trapezoidal rule, the approximate value of the integral \(\int_{0}^{6} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\) is

1 1.3128
2 1.4108
3 1.4218
4 None of these
CG PET- 2015
Differential Equation

87618 Using trapezoidal rule and taking \(n=4\), the approximate value of integral \(\int_{1}^{9} x^{2} d x\) is
\(2\left[\frac{1}{2}\left(1+9^{2}\right)+\alpha^{2}+\beta^{2}+7^{2}\right]\) then

1 \(\alpha=1, \beta=3\)
2 \(\alpha=2, \beta=4\)
3 \(\alpha=3, \beta=5\)
4 \(\alpha=4, \beta=6\)
CG PET- 2018
Differential Equation

87611 The positive root of equation \(x^{3}-2 x-5=0\) lies in the interval

1 \((0,1)\)
2 \((1,2)\)
3 \((2,3)\)
4 \((3,4)\)
CG PET- 2016
Differential Equation

87612 The value of \(\int_{0}^{6} \frac{d x}{1+x^{2}}\) by choosing six subintervals and by using Simpson's Rule will be

1 1.3562
2 1.3662
3 1.3456
4 1.2662
CG PET- 2016