BCA Semester 1 – Mathematics I () Questions & Answers | Past TU Exam Papers

Given points: \( P(1, -1, 0) \), \( Q(2, 1, -1) \), \( R(-1, 1, 2) \) Direction vectors in the plane: \( \vec{PQ} = Q - P = (2 - 1,\ 1 - (-1),\ -1 - 0) = (1,\ 2,\ -1) \) \( \vec{PR} = R - P = (-1 - 1,

Given word: "Sunday" The word "Sunday" has \( 6 \) different letters: \( S, U, N, D, A, Y \) Total number of arrangements: \[ = 6! = 720 \] a) Number of ways the letters of the word "Sunday" can be ar

$\textbf{Given ellipse:} $ \[ 4x^2 + 9y^2 = 36 \] $\textbf{Solution:}$ $\textbf{Step 1:}$ Write the equation in standard form: \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] Here, $a^2 = 9 \implies a =

$\textbf{Given:} $ $\vec{a}$ and $\vec{b}$ are unit vectors, and $\theta$ is the angle between them. $\textbf{To prove:}$ \[ |\vec{a} - \vec{b}| = \sin \frac{\theta}{2} \] $\textbf{Solution:}$

$\textbf{To prove by vector method:} $ \[ \cos(A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B \] $\textbf{Solution:} $ $\textbf{Step 1:}$ Consider two unit vectors $\vec{u}$ and $\vec{v}$ making

$\textbf{Given points:}$ $(1,2)$, $(3,1)$, $(-3,-1)$ $\textbf{Solution:}$ Step 1: General equation of a circle: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] Step 2: Substitute the points into the equatio

$\textbf{Given:}$ $\text{Latus rectum } l = 3, \quad \text{eccentricity } e = \frac{1}{\sqrt{2}}$ $\textbf{Solution:}$ $\textbf{Step 1}$: Recall formula for latus rectum of an ellipse: \[ l = \f

$\textbf{To prove:}$ \[ \cos(A - B) = \cos A \cos B + \sin A \sin B \] $\textbf{Solution:}$ $\textbf{Step 1:}$ Consider two unit vectors $\vec{u}$ and $\vec{v}$ making angles $A$ and $B$ with the

$\textbf{Problem a: Find the angle between vectors }$ \[ \mathbf{u} = 4\mathbf{i} - 2\mathbf{j} + \mathbf{k}, \quad \mathbf{v} = \mathbf{i} + \mathbf{j} - \mathbf{k} \] $\textbf{Step 1:}$ Recall form

$\textbf{Given:}$ \[ 25x^2 + 4y^2 = 100 \] $\textbf{Step 1:}$ Write in standard form of ellipse: \[ \frac{x^2}{4} + \frac{y^2}{25} = 1 \] $\text{Thus, } a^2 = 25, \quad b^2 = 4$ (since $y$-axis

$\textbf{Given:}$ Vertex $(h, k) = (?, 2)$ and directrix $x = 4$. $\textbf{Step 1:}$ Recall the standard equation of a parabola with a vertical axis: \[ (y - k)^2 = 4p(x - h) \] where $(h, k)$ i

$\textbf{Problem a:}$ Define a parabola and derive standard equation $y^2 = 4ax$ $\textbf{Definition:}$ A parabola is the locus of a point which moves such that its distance from a fixed point (focus)

$\textbf{Given:}$ Latus rectum $l = 5$, eccentricity $e = \frac{1}{\sqrt{2}}$. We need the equation of the ellipse. $\textbf{Step 1:}$ Standard form of ellipse along x-axis: \[ \frac{x^2}{a^2} + \fr

$\textbf{Given:}$ \[ \vec a = \sqrt{3} \, \hat{i} + \hat{j}, \quad \vec a \times \vec b = (1, 2, 2) \] $\text{Find the angle } \theta \text{ between } \vec a \text{ and } \vec b.$ $\textbf{Step 1:

$\textbf{Problem a:}$ Find the equation of a hyperbola with focus $F(-2,0)$ and directrix $x = -\frac{1}{2}$. $\textbf{Step 1:}$ Recall definition of a hyperbola: $\text{Distance of any point } P(x,