Step 1: Understanding the function.
The function is defined as \( f(x) = x^2 + 1 \), where \( x \in \mathbb{R} \). The term \( x^2 \) is always non-negative, and adding 1 shifts the range of \( x^2 \) to start from 1.
Step 2: Range of \( f(x) \).
The minimum value of \( x^2 \) is 0, so the minimum value of \( f(x) = x^2 + 1 \) is: \[ f(x) = 0 + 1 = 1. \] Thus, the range of \( f(x) \) is \( [1, \infty) \).
Step 3: Conditions for \( f(x) \) to be onto.
For \( f(x) \) to be onto, the codomain \( A \) must include the entire range of \( f(x) \). Therefore, \( A = [1, \infty) \).
Step 4: Conclusion.
The function \( f(x) = x^2 + 1 \) is onto if \( A = [1, \infty) \). {10pt}
Show that \( R \) is an equivalence relation. Also, write the equivalence class \([2]\).
List-I | List-II |
(A) Absolute maximum value | (I) 3 |
(B) Absolute minimum value | (II) 0 |
(C) Point of maxima | (III) -5 |
(D) Point of minima | (IV) 4 |
In number theory, it is often important to find factors of an integer \( N \). The number \( N \) has two trivial factors, namely 1 and \( N \). Any other factor, if it exists, is called a non-trivial factor of \( N \). Naresh has plotted a graph of some constraints (linear inequations) with points \( A(0, 50) \), \( B(20, 40) \), \( C(50, 100) \), \( D(0, 200) \), and \( E(100, 0) \). This graph is constructed using three non-trivial constraints and two trivial constraints. One of the non-trivial constraints is \( x + 2y \geq 100 \).
Based on the above information, answer the following questions:
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.