87605
If \(\mathrm{e}^{0}=1, \mathrm{e}^{1}=2.72, \mathrm{e}^{2}=7.39, \mathrm{e}^{3}=20.09, \mathrm{e}^{4}=54.60\) then the value of \(\int_{0}^{4} e^{x} d x\) using simpson's rule, will be
87606
The value of \(\sqrt{12}\) upto three places of decimals using the method of Newton-Raphson, will be
1 3.463
2 3.462
3 3.467
4 None of these
Explanation:
(A) : Let's consider \(x=\sqrt{12}\) or \(x^{2}=12\) On differentiating w.r.t \(x\), we get \(f^{\prime}(x)=2 x\) \(\because \quad \mathrm{f}(3)\lt 0\) and \(\mathrm{f}(4)>0\) Hence, root will lie between 3 and 4 \(|\mathrm{f}(3)|\lt |\mathrm{f}(4)|\) \(\therefore \quad \mathrm{x}_{0}=3\) First iteration, \(\mathrm{x}_{1}=\mathrm{x}_{0}-\frac{\mathrm{f}\left(\mathrm{x}_{0}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)}=3-\frac{(9-12)}{2 \times 3}=3+\frac{3}{6}=3.5\) Now, second iteration, \(\mathrm{x}_{2}=3.5-\frac{\mathrm{f}(3.5)}{\mathrm{f}^{\prime}(3.5)}=3.5-\frac{\left[(3.5)^{2}-12\right]}{2 \times 3.5}\) \(=3.5-\frac{[12.25-12]}{7}=\frac{24.5-0.25}{7.0}=3.463\)
CG PET- 2014
Differential Equation
87607
According to Simpson's rule, the value of \(\int_{1}^{7} \frac{d x}{x}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87605
If \(\mathrm{e}^{0}=1, \mathrm{e}^{1}=2.72, \mathrm{e}^{2}=7.39, \mathrm{e}^{3}=20.09, \mathrm{e}^{4}=54.60\) then the value of \(\int_{0}^{4} e^{x} d x\) using simpson's rule, will be
87606
The value of \(\sqrt{12}\) upto three places of decimals using the method of Newton-Raphson, will be
1 3.463
2 3.462
3 3.467
4 None of these
Explanation:
(A) : Let's consider \(x=\sqrt{12}\) or \(x^{2}=12\) On differentiating w.r.t \(x\), we get \(f^{\prime}(x)=2 x\) \(\because \quad \mathrm{f}(3)\lt 0\) and \(\mathrm{f}(4)>0\) Hence, root will lie between 3 and 4 \(|\mathrm{f}(3)|\lt |\mathrm{f}(4)|\) \(\therefore \quad \mathrm{x}_{0}=3\) First iteration, \(\mathrm{x}_{1}=\mathrm{x}_{0}-\frac{\mathrm{f}\left(\mathrm{x}_{0}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)}=3-\frac{(9-12)}{2 \times 3}=3+\frac{3}{6}=3.5\) Now, second iteration, \(\mathrm{x}_{2}=3.5-\frac{\mathrm{f}(3.5)}{\mathrm{f}^{\prime}(3.5)}=3.5-\frac{\left[(3.5)^{2}-12\right]}{2 \times 3.5}\) \(=3.5-\frac{[12.25-12]}{7}=\frac{24.5-0.25}{7.0}=3.463\)
CG PET- 2014
Differential Equation
87607
According to Simpson's rule, the value of \(\int_{1}^{7} \frac{d x}{x}\) is
87605
If \(\mathrm{e}^{0}=1, \mathrm{e}^{1}=2.72, \mathrm{e}^{2}=7.39, \mathrm{e}^{3}=20.09, \mathrm{e}^{4}=54.60\) then the value of \(\int_{0}^{4} e^{x} d x\) using simpson's rule, will be
87606
The value of \(\sqrt{12}\) upto three places of decimals using the method of Newton-Raphson, will be
1 3.463
2 3.462
3 3.467
4 None of these
Explanation:
(A) : Let's consider \(x=\sqrt{12}\) or \(x^{2}=12\) On differentiating w.r.t \(x\), we get \(f^{\prime}(x)=2 x\) \(\because \quad \mathrm{f}(3)\lt 0\) and \(\mathrm{f}(4)>0\) Hence, root will lie between 3 and 4 \(|\mathrm{f}(3)|\lt |\mathrm{f}(4)|\) \(\therefore \quad \mathrm{x}_{0}=3\) First iteration, \(\mathrm{x}_{1}=\mathrm{x}_{0}-\frac{\mathrm{f}\left(\mathrm{x}_{0}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)}=3-\frac{(9-12)}{2 \times 3}=3+\frac{3}{6}=3.5\) Now, second iteration, \(\mathrm{x}_{2}=3.5-\frac{\mathrm{f}(3.5)}{\mathrm{f}^{\prime}(3.5)}=3.5-\frac{\left[(3.5)^{2}-12\right]}{2 \times 3.5}\) \(=3.5-\frac{[12.25-12]}{7}=\frac{24.5-0.25}{7.0}=3.463\)
CG PET- 2014
Differential Equation
87607
According to Simpson's rule, the value of \(\int_{1}^{7} \frac{d x}{x}\) is
87605
If \(\mathrm{e}^{0}=1, \mathrm{e}^{1}=2.72, \mathrm{e}^{2}=7.39, \mathrm{e}^{3}=20.09, \mathrm{e}^{4}=54.60\) then the value of \(\int_{0}^{4} e^{x} d x\) using simpson's rule, will be
87606
The value of \(\sqrt{12}\) upto three places of decimals using the method of Newton-Raphson, will be
1 3.463
2 3.462
3 3.467
4 None of these
Explanation:
(A) : Let's consider \(x=\sqrt{12}\) or \(x^{2}=12\) On differentiating w.r.t \(x\), we get \(f^{\prime}(x)=2 x\) \(\because \quad \mathrm{f}(3)\lt 0\) and \(\mathrm{f}(4)>0\) Hence, root will lie between 3 and 4 \(|\mathrm{f}(3)|\lt |\mathrm{f}(4)|\) \(\therefore \quad \mathrm{x}_{0}=3\) First iteration, \(\mathrm{x}_{1}=\mathrm{x}_{0}-\frac{\mathrm{f}\left(\mathrm{x}_{0}\right)}{\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)}=3-\frac{(9-12)}{2 \times 3}=3+\frac{3}{6}=3.5\) Now, second iteration, \(\mathrm{x}_{2}=3.5-\frac{\mathrm{f}(3.5)}{\mathrm{f}^{\prime}(3.5)}=3.5-\frac{\left[(3.5)^{2}-12\right]}{2 \times 3.5}\) \(=3.5-\frac{[12.25-12]}{7}=\frac{24.5-0.25}{7.0}=3.463\)
CG PET- 2014
Differential Equation
87607
According to Simpson's rule, the value of \(\int_{1}^{7} \frac{d x}{x}\) is
