List of top Engineering Mathematics Questions asked in GATE Aerospace Engineering

\( \hat{i} \) and \( \hat{j} \) denote unit vectors in the \( x \) and \( y \) directions, respectively. The outward flux of the two-dimensional vector field \( \vec{v} = x \hat{i} + y \hat{j} \) over the unit circle centered at the origin is ___________ (rounded off to two decimal places).

An approximate solution of the equation \( x^3 - 17 = 0 \) is to be obtained using the Newton-Raphson method. If the initial guess is \( x_0 = 2 \), the value at the end of the first iteration is \( x_1 = \)  ________ (rounded off to two decimal places).

Consider the ordinary differential equation:

The equation of a closed curve in two-dimensional polar coordinates is given by \( r = \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \). The area enclosed by the curve is ___________ (answer in integer).

The maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain [0, 3] occurs at \( x = \) _________ (rounded off to two decimal places).

Let the function \( f(x) \) be defined as:
\[ f(x) = \begin{cases} A + x, & \text{if } x < 2 \\ 1 + x^2, & \text{if } x \geq 2 \end{cases} \] If the function \( f(x) \) is continuous at \( x = 2 \), the value of \( A \) should be _______.

The partial differential equation \[ \frac{\partial^2 u}{\partial x^2} + 4 \frac{\partial^2 u}{\partial x \partial y} + 2 \frac{\partial^2 u}{\partial y^2} = 0 \] is ________.

An approximate solution of the equation \( x^3 - 17 = 0 \) is to be obtained using the Newton-Raphson method. If the initial guess is \( x_0 = 2 \), the value at the end of the first iteration is \( x_1 = \) \_\_\_\_\_\ (rounded off to two decimal places).
Consider the ordinary differential equation: \[ \frac{1}{2} \frac{dy}{dx} + \frac{y}{x} = 1. { If } y = \frac{2}{3} { at } x = 1, { then the value of } y { at } x = 3 { is } \_\_\_\_\_\_ { (rounded off to the nearest integer).} \]
\( \hat{i} \) and \( \hat{j} \) denote unit vectors in the \( x \) and \( y \) directions, respectively. The outward flux of the two-dimensional vector field \( \vec{v} = x \hat{i} + y \hat{j} \) over the unit circle centered at the origin is \_\_\_\_\_\_ (rounded off to two decimal places).
The equation of a closed curve in two-dimensional polar coordinates is given by \( r = \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \). The area enclosed by the curve is \_\_\_\_\_\_\_\_ (answer in integer).
Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]
The maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain [0, 3] occurs at \( x = \) \_\_\_\_\_\_ (rounded off to two decimal places).
The partial differential equation \[ \frac{\partial^2 u}{\partial x^2} + 4 \frac{\partial^2 u}{\partial x \partial y} + 2 \frac{\partial^2 u}{\partial y^2} = 0 \] is \_\_\_\_\_\_.
Let the function \( f(x) \) be defined as: \[ f(x) = \begin{cases} A + x, & {if } x<2
1 + x^2, & {if } x \geq 2 \end{cases} \] If the function \( f(x) \) is continuous at \( x = 2 \), the value of \( A \) is \_\_\_\_\_.
If \( i = \sqrt{-1} \), \[ \frac{(i+1)^3}{i-1} = \_\_\_\_\_\_\_\_. \]
Consider the ordinary differential equation: \[ \frac{1}{2} \frac{dy}{dx} + \frac{y}{x} = 1. { If } y = \frac{2}{3} { at } x = 1, { then the value of } y { at } x = 3 { is } \_\_\_\_\_\_ { (rounded off to the nearest integer).} \]
\( \hat{i} \) and \( \hat{j} \) denote unit vectors in the \( x \) and \( y \) directions, respectively. The outward flux of the two-dimensional vector field \( \vec{v} = x \hat{i} + y \hat{j} \) over the unit circle centered at the origin is \_\_\_\_\_\_ (rounded off to two decimal places).
The equation of a closed curve in two-dimensional polar coordinates is given by \( r = \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \). The area enclosed by the curve is \_\_\_\_\_\_\_\_ (answer in integer).
An approximate solution of the equation \( x^3 - 17 = 0 \) is to be obtained using the Newton-Raphson method. If the initial guess is \( x_0 = 2 \), the value at the end of the first iteration is \( x_1 = \) \_\_\_\_\_\ (rounded off to two decimal places).
The maximum value of the function \( f(x) = (x - 1)(x - 2)(x - 3) \) in the domain [0, 3] occurs at \( x = \) \_\_\_\_\_\_ (rounded off to two decimal places).
Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]
The partial differential equation \[ \frac{\partial^2 u}{\partial x^2} + 4 \frac{\partial^2 u}{\partial x \partial y} + 2 \frac{\partial^2 u}{\partial y^2} = 0 \] is \_\_\_\_\_\_.
If \( i = \sqrt{-1} \), \[ \frac{(i+1)^3}{i-1} = \_\_\_\_\_\_\_\_. \]
Let the function \( f(x) \) be defined as: \[ f(x) = \begin{cases} A + x, & {if } x<2
1 + x^2, & {if } x \geq 2 \end{cases} \] If the function \( f(x) \) is continuous at \( x = 2 \), the value of \( A \) is \_\_\_\_\_.