GATE Overflow for GATE IN - Questions without answers in Numerical Methods

GATE IN 2025 | GA | Question: 8

Thu, 06 Mar 2025 17:22:57 +0000

Which one of the following options is correct for the given data in the table?

$\begin{array}{|c|c|c|c|c|}\hline\text{Iteration (i)} & 0 & 1 & 2 & 3 \\ \hline\text{Input (I)} & 20 & -4 & 10 & 15 \\ \hline text{Output } X & 20 & 16 & 26 & 41 \\ \hline \text{Output } Y & 20 & -80 & -800 & -12000 \\
\hline \end{array}$

$X(i)=X(i-1)+I(i) ; \quad Y(i)=Y(i-1) I(i) ; \quad i>0$
$X(i)=X(i-1) I(i) ; \quad Y(i)=Y(i-1)+I(i) ; \quad i>0$
$X(i)=X(i-1) I(i) ; \quad Y(i)=Y(i-1) I(i) ; \quad i>0$
$X(i)=X(i-1)+I(i) ; \quad Y(i)=Y(i-1) I(i-1) ; \quad i>0$

GATE IN 2025 | Question: 27

Thu, 06 Mar 2025 17:18:01 +0000

Newton-Raphson method is used to compute the inverse of the number $1.6$. Among the following options, the initial guess of the solution that results in non-convergence of the iterative process is

$0.55$
$0.75$
$1.15$
$1.25$

GATE IN 2024 | Question: 24

Fri, 16 Feb 2024 18:41:00 +0000

Consider a system given by the following first order differential equation:

$$ \frac{d y}{d t}=y+2 t-t^{2} $$

where, $y(0)=1$ and $0 \leq t<\infty$. Using a step size $h=0.1$ for the improved Euler method, the value of $y(t)$ at $t=0.1$ is $\_\_\_\_\_\_$ (rounded off to two decimal places).

GATE IN 2022 | Question: 24

Sun, 20 Mar 2022 17:33:58 +0000

The Newton-Raphson method is applied to determine the solution of $f(x) = 0$ where $f(x) = x – \cos (x).$ If the initial guess of the solution is $x_{0} = 0,$ the value of the next approximation $x_{1}$ is ________ (round off to two decimal places)

GATE IN 2017 | Question: 28

Mon, 02 Nov 2020 00:00:21 +0000

The following table lists an $n^{th}$ order polynominal $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ and the forward differences evaluated at equally spaced values of $x$. The order of the polynominal is

GATE2014-27

Sun, 25 Mar 2018 23:10:48 +0000

The iteration step in order to solve for the cube roots of a given number N using the Newton-Raphson’s method is

$x_{k+1}=x_k+\frac{1}{3}(N-x^3_k)$
$x_{k+1}=\frac{1}{3}(2x_k+\frac{N}{x^2_k})$
$x_{k+1}=x_k-\frac{1}{3}(N-x^3_k)$
$x_{k+1}=\frac{1}{3}(2x_k-\frac{N}{x^2_k})$

GATE IN 2013 | Question: 26

Sun, 25 Mar 2018 19:30:32 +0000

While numerically solving the differential equation $\frac{dy}{dx}+2xy^2=0,\; y(0)=1$ using Euler’s predictor-corrector (improved Euler-Cauchy) method with a step size of 0.2, the value of $y$ after the first step is

$1.00$
$1.03$
$0.97$
$0.96$

GATE IN 2012 | Question: 5

Sun, 25 Mar 2018 09:53:01 +0000

Given $f(z)=\frac{1}{z+1}-\frac{2}{z+3}.$ If $C$ is a counterclockwise path in the $z$-plane such that $|z+1|=1,$ the value of $\frac{1}{2\pi j}\oint_cf(z)dz$ is

$-2$
$-1$
$1$
$2$

GATE2018IN: 37

Tue, 20 Feb 2018 12:03:45 +0000

Consider the linear system x = $\begin{bmatrix}-1 & 0 \\0 & -2\end{bmatrix} x,$ with initial condition $x(0) = \begin{bmatrix}1 \\1\end{bmatrix}$. The solution $x(t)$ for this system is

$x(t) = \begin{bmatrix}e^{-t} & te^{-2t} \\0 & e^{-2t}\end{bmatrix}$ $\begin{bmatrix}1 \\1\end{bmatrix}$
$x(t) = \begin{bmatrix}e^{-t} & 0 \\0 & e^{2t}\end{bmatrix}$ $\begin{bmatrix}1 \\1\end{bmatrix}$
$x(t) = \begin{bmatrix}e^{-t} & -t^2e^{-2t} \\0 & e^{-2t}\end{bmatrix}$ $\begin{bmatrix}1 \\1\end{bmatrix}$
$x(t) = \begin{bmatrix}e^{-t} & 0 \\0 & e^{-2t}\end{bmatrix}$ $\begin{bmatrix}1 \\1\end{bmatrix}$