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

Given function: \( f(x) = \dfrac{2x + 1}{3 - x} \) Domain: The denominator must not be equal to \( 0 \). \( 3 - x \neq 0 \) \( x \neq 3 \) So, the domain is: All real numbers except \( x = 3 \). Range

$\textbf{Given:} $ \[ f: \mathbb{R} \to \mathbb{R},\ f(x) = 2x + 3 \] \[ g: \mathbb{R} \to \mathbb{R},\ g(x) = x^2 - 1 \] $\textbf{Solution:}$ $\textbf{i) }$ $f \circ g$ \[ \quad (f \circ g)(x)

$\textbf{Definition:}$ $\text{A function } f \text{ from a set } A \text{ to a set } B \text{ is a rule that assigns each element of } A \text{ exactly one element of } B.$ --- $\textbf{Domain:}$

$\textbf{Definition of a function:}$ $\text{A function } f \text{ from a set } A \text{ to a set } B \text{ is a rule that assigns each element of } A \text{ exactly one element of } B.$ --- $\tex

$\textbf{Definition:}$ $\text{Exponential function: } f(x) = a^x \text{ (with } a > 0, a \neq 1\text{) assigns each real number } x \text{ to } a^x.$ $\text{Logarithmic function: } f(x) = \loga x

$\textbf{Given function:}$ \[ f(x) = \sqrt{6 - x - x^2} \] $\textbf{Step 1: Determine the domain}$ $\text{Since square root is defined for non-negative values:}$ \[ 6 - x - x^2 \ge 0 \implies -x

$\textbf{Given function:}$ \[ f: \mathbb{R} \to \mathbb{R}, \quad f(x) = 3x - 1 \] $\textbf{Step 1:}$ Show injective (one-to-one) $\text{Assume } f(x1) = f(x2) \implies 3x1 - 1 = 3x2 - 1 \implies 3x

$\textbf{Given function:}$ \[ f(x) = \sqrt{6 - x - x^2} \] $\textbf{Step 1: Determine the domain}$ $\text{Since the square root is defined for non-negative values:}$ \[ 6 - x - x^2 \ge 0 \implie