Numerical Methods

Question 1
Let y^{2}-2y+1=x and \sqrt{x}+y=5. The value of x+\sqrt{y} equals _________. (Give the answer up to three decimal places)
A
3.732
B
4.734
C
5.734
D
6.732
GATE EE 2017-SET-2   Engineering Mathematics
Question 1 Explanation: 
\begin{aligned} \sqrt{x}+y&=5\\ \sqrt{x}+1+5x&=5\\ 2\sqrt{x}&=4\\ x&=4\\ y^2-2y+1&=x\\ (y-1)^2&=x\\ y-1&=\sqrt{x}\\ y&=1+\sqrt{x}\\ y&=1+\sqrt{4}=3\\ \therefore \;\;x+\sqrt{y}&=4+\sqrt{3}=5.732 \end{aligned}
Question 2
Let x and y be integers satisfying the following equations
2x^{2}+y^{2}=34
x+2y=11
The value of (x+y) is ________.
A
4
B
3
C
7
D
10
GATE EE 2017-SET-2   Engineering Mathematics
Question 2 Explanation: 
\begin{aligned} x+2y=1\\ x=11-2y\\ 2x^2+y^2=34\\ 2(11-2y)^2+y^2=34\\ 24^2+8y^2-88y+y^2-34=0\\ 9y^2-88y+208=0\\ y=5.77,4\\ x=-0.54\\ x=11-2(4)=3\\ x=3,y=4\\ x+y=7 \end{aligned}
Question 3
Only one of the real roots of f(x)=x^{6}-x-1 lies in the interval 1\leq x\leq 2 and bisection method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is ________.
A
1000
B
100
C
10
D
1
GATE EE 2017-SET-1   Engineering Mathematics
Question 3 Explanation: 
\begin{aligned} f(x)&=x^6-x-1\\ a&=1, b=2, \in =0.01=10^{-3} \end{aligned}
The minimum number of iterations by Bisection method is given by
\begin{aligned} \frac{|b-a|}{2^n} &\lt \in\\ \frac{2-1}{2^n} &\lt 10^{-3}\\ \frac{1}{2^n} &\lt \frac{1}{10^{-3}}\\ 2^n &\gt 10^3\\ n &\gt \frac{\ln 1000}{\ln 2}\\ n &\gt 9.96\\ n&=10 \end{aligned}
Question 4
A differential equation \frac{di}{dt}-0.2i=0 is applicable over -10\lt t \lt 10. If i(4)=10, then i(-5) is _____.
A
1
B
0.5
C
1.6
D
2.4
GATE EE 2015-SET-2   Engineering Mathematics
Question 4 Explanation: 
\begin{aligned} \frac{di}{dt} &=0.2i \\ \frac{di}{i}&=0.2 \; dt \\ \int \frac{di}{i} &=\int 0.2 \; dt \\ \log i &=0.2t+ \log C \\ \log i-\log C &=0.2t \\ \log \left ( \frac{i}{C} \right ) &= 0.2t\\ \frac{i}{C} &= e^{0.2t}\\ i &=C e^{0.2t} \\ i(4)=10\;\;i.e.\;\;i &=10 \text{ when } t=4 \\ 10&=Ce^{(0.2)4} \\ 10&= C(2.225)\\ C&=4.493 \\ i&=(4.493)e^{0.2t} \\ \text{ when, } i&=-5 \\ i&=(4.493)e^{(0.2)(-5)}\\ &=1.652 \end{aligned}
Question 5
Two coins R and S are tossed. The 4 joint events H_{R}H_{S},T_{R}T_{S},h_{R}H_{S},T_{R}H_{S} have probabilities 0.28, 0.18, 0.30, 0.24, respectively, where H represents head and T represents tail. Which one of the following is TRUE?
A
The coin tosses are independent.
B
R is fair, S is not.
C
S is fair, R is not
D
The coin tosses are dependent
GATE EE 2015-SET-2   Engineering Mathematics
Question 5 Explanation: 
From the given information, we can create a joint probability table as follows:

From the table, we can get
P(H_R)=0.58, P(T_R)=0.42, P(H_S)=0.52, P(T_S)=0.48
So, Coins R and S are biased (not fair). So choises (B) and (C) are both false.
The coin tosses are not independent since their probability of heads and tails is not 0.5.
R and S are dependent.
If R and S were independent then all the joint probabilities will be equal to the product of the marginal probabilities.
For example
P(H_R\cap H_S)=0.28
P(H_R)\cdot P(H_S)=0.58 \times 0.52=0.3016
Clearly P(H_R\cap H_S) \neq P(H_R)\cdot P(H_S)
So R and S are not independent.
i.e. R and S are dependent. So, Choise (A) is also false and choise (D) is true.
Question 6
The function f(x)=e^{x}-1 is to be solved using Newton-Raphson method. If the initial value of x_0 is taken as 1.0, then the absolute error observed at 2^{nd} iteration is _____.
A
0.01
B
0.03
C
0.06
D
0.09
GATE EE 2014-SET-3   Engineering Mathematics
Question 6 Explanation: 
Given, f(x)=e^x-1
or, f(x_K)=e^{x_K}-1 and x_0=1
In Newton-Raphson method, we have
\begin{aligned} x_{K+1}&=\left [ x_K-\frac{f(x_K)}{f'(x_K)} \right ] \\ x_1&= \left [ x_0- \frac{f(x_0)}{f'(x_0)}\right ]\;\;...(i)\\ \text{Now, } f(x_0)&=e^{x_0}-1=e^1-1=(e-1) \\ \text{and, } f'(x)&=e6x \\ \therefore \;\; f'(x_0) &=e^1=e \end{aligned}
Putting the values, we get:
\begin{aligned} x_1&=\left [ 1-\frac{e-1}{e} \right ]=\left [ \frac{e-e+1}{e} \right ]=e^{-1}\\ \text{Also, }x_2&=\left [ x_1-\frac{f(x_1)}{f'(x_1)} \right ]\;\;...(ii)\\ \text{Now, }x_1&=\frac{1}{e}=e^{-1} \text{ and } f(x_1)=(e^{e^{-1}}-1)\\ f'(x_1)&=e^{e^{-1}} \end{aligned}
Putting the value, we get:
\begin{aligned} x_2&=\left [ e^{-1}-\frac{(e^{e^{-1}}-1)}{e^{e^{-1}}} \right ]\\ &=\left [ e^{-1}-\frac{(e^{0.37}-1)}{e^{0.37}} \right ]=0.06 \end{aligned}
Question 7
When the Newton-Raphson method is applied to solve the equation f(x)=x^{3}+2x-1=0, the solution at the end of the first iteration with the initial value as x_{0}=1.2 is
A
-0.82
B
0.49
C
0.705
D
1.69
GATE EE 2013   Engineering Mathematics
Question 7 Explanation: 
\begin{aligned} f'(x)&=3x^2+2\\ f'(x_0)&=3(1.2)^2+2=6.32\\ f(x_0)&=(1.2)^3+2 \times 1.2-1=3.128\\ f'(x_1)&=x_0-\frac{f(x_0)}{f'(x_1)}=1.2-\frac{3.728}{6.32}=0.705 \end{aligned}
Question 8
Solution of the variables x_1 \; and \; x_2 for the following equations is to be obtained by employing the Newton-Raphson iterative method.
equation (1) 10x_{2}sin x_{1}-0.8=0
equation (2) 10x_{2}^{2}-10 x_{2} cos x_{1}-0.6=0
Assuming the initial values are x_1=0.0 \; and \; x_2=1.0, the jacobian matrix is
A
\begin{bmatrix} 10 & -0.8\\ 0 &-0.6 \end{bmatrix}
B
\begin{bmatrix} 10 & 0\\ 0 &10 \end{bmatrix}
C
\begin{bmatrix} 0 & -0.8\\ 10 &-0.6 \end{bmatrix}
D
\begin{bmatrix} 10 & 0\\ 10 &-10 \end{bmatrix}
GATE EE 2011   Engineering Mathematics
Question 8 Explanation: 
\begin{aligned} u(x_1,x_2)&=10x_2 \sin x_1-0.8=0\\ v(x_1,x_2)&=10x_2^2-10x_2 \cos x_1-0.6=0\\ &\text{The Jacobian matrix is }\\ \begin{bmatrix} \frac{\partial u}{\partial x_1} & \frac{\partial u}{\partial x_2} \\ \frac{\partial v}{\partial x_1} & \frac{\partial v}{\partial x_2} \end{bmatrix}&=\begin{bmatrix} 10x_2 \cos x_1 & 10 \sin x_1\\ 10x_2 \sin x_1& 20x_2-10 \cos x_1 \end{bmatrix}\\ &=\begin{bmatrix} 10 & 0\\ 0& 10 \end{bmatrix} \end{aligned}
Question 9
Let x^{2}- 117 = 0. The iterative steps for the solution using Newton-Raphon's method is given by
A
x_{k+1}=\frac{1}{2}(x_{k}+\frac{117}{x_{k}})
B
x_{k+1}=(x_{k}-\frac{117}{x_{k}})
C
x_{k+1}=(x_{k}-\frac{x_{k}}{117})
D
x_{k+1}=x_{k}-\frac{1}{2}(x_{k}+\frac{117}{x_{k}})
GATE EE 2009   Engineering Mathematics
Question 9 Explanation: 
\begin{aligned} x_{k+1}&=x_k-\frac{f(x_k)}{f'(x_k)}\\ &=x_k-\frac{x_k^2-117}{2x_k}\\ &=\frac{1}{2}\left [ x_k+\frac{117}{x_k} \right ] \end{aligned}
Question 10
Equation e^{x} - 1 = 0 is required to be solved using Newton's method with an initial guess x_0=-1. Then, after one step of Newton's method, estimate x_1 of the solution will be given by
A
0.71828
B
0.36784
C
0.20587
D
0
GATE EE 2008   Engineering Mathematics
Question 10 Explanation: 
Here, \begin{aligned} f(x)&=e^x-1\\ f(x)&=e^x \end{aligned}
The newton raphson iterative equation is
\begin{aligned} x_{i+1}&=x_i-\frac{f(x_i)}{f'(x_i)}\\ f(x_i)&=e^{xi}-1\\ f'(x_i)&=e^{xi}\\ \therefore \;\; x_{i+1}&=x_i-\frac{e^{x_i}-1}{e^{x_i}}\\ i.e.\;\;x_{i+1}&=\frac{x_ie^{x_i}-(e^{x_i}-1)}{e^{x_i}}\\ &=\frac{e^{x_i}(x_i-1)+1}{e^{x_i}}\\ \text{Now put }i&=0\\ x_1&=\frac{e^{x_0}(x_0-1)+1}{e^{x_0}}\\ \text{put, }x_0&=-1 \text{ as given,}\\ x_1&=[e^{-1}(-2)+1]/e^{-1}\\ &=0.71828 \end{aligned}
There are 10 questions to complete.
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