Relations and Functions: Class 11 Miscellaneous Exercise Complete Guide

Core Concepts: Analyzing Real-Valued Functions

Real-valued functions are the bedrock of higher mathematics. In the relations and functions miscellaneous ex class 11 ncert, you are frequently tasked with identifying the domain and range of tricky algebraic functions. To determine the domain of a real function $y = f(x)$, we must identify all real values of $x$ for which $f(x)$ is defined and yields a real number. Two primary algebraic constraints dictate this process: first, the denominator of any rational fraction must never equal zero, as division by zero is mathematically undefined. Second, the expression under a square root (or any even root) must be non-negative ($f(x) \ge 0$) to avoid imaginary numbers. Solving inequalities using interval notation is essential. Finding the range requires setting $y = f(x)$, expressing $x$ in terms of $y$, and applying real-number constraints on $y$ to ensure we only include values that the function can actually output.

Key Frameworks & Terminology

Real Function

A function whose domain and codomain are both subsets of the set of real numbers (R).

Algebra of Real Functions

Operations defining addition, subtraction, multiplication, and division of real functions over the intersection of their domains.

Modulus Function

The function f(x) = |x|, which outputs the non-negative absolute value of any real input x.

Step-by-Step Strategy to Find Domain and Range

1. Step 1: Set Up Algebraic Constraints — Look at the function structure. If there is a square root, set the term under the radical to be $\ge 0$. If there is a denominator, set it $\ne 0$.
2. Step 2: Solve the Inequalities — Solve the algebraic inequalities to find the set of valid input values. Express this set using proper interval notation to write down the Domain.
3. Step 3: Express x in Terms of y — Set $y = f(x)$. Rearrange the algebraic equation to isolate $x$, writing it as a function of $y$ (i.e., $x = g(y)$).
4. Step 4: Determine the Range — Apply constraints on $y$ so that $x$ remains real and within the original domain. This set of valid $y$ values forms the Range.

Critical Exam Traps & Boards Guidelines

1. Strict vs. Non-strict Inequalities: When solving for the domain of $1/\sqrt{g(x)}$, remember that $g(x)$ must be strictly greater than zero ($g(x) > 0$). It cannot equal zero because it sits in the denominator!
2. The Modulus Trap: Remember that $|x-a|$ is always non-negative. If $f(x) = -|x-1|$, the output values are always less than or equal to zero, making the range $(-\infty, 0]$.
3. Domain Intersection: When combining two functions, such as $f+g$ or $f/g$, the domain of the resulting function is the intersection of their individual domains ($D_f \cap D_g$), excluding any points where the denominator of the quotient becomes zero.

Practice Questions with Solutions

• Q: Find the domain of the function $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$. A: Step 1: Identify the constraint. Since this is a rational function, the denominator must not be zero. Step 2: Set the denominator to zero and solve: $x^2 - 8x + 12 = 0$. Step 3: Factorize the quadratic equation: $(x-2)(x-6) = 0$, which gives $x = 2$ and $x = 6$. Step 4: Exclude these points from the set of real numbers. Final answer: The domain of the function is $\mathbb{R} - \{2, 6\}$.
• Q: Find the domain and range of the real function $f(x) = \sqrt{9 - x^2}$. A: Step 1: Set up the radical constraint. The term under the square root must be non-negative: $9 - x^2 \ge 0$. Step 2: Solve the inequality: $x^2 \le 9 \implies -3 \le x \le 3$. This gives the Domain: $[-3, 3]$. Step 3: Set $y = \sqrt{9-x^2}$. Since $y$ is a principal square root, $y \ge 0$. Step 4: Square both sides to express $x$ in terms of $y$: $y^2 = 9 - x^2 \implies x^2 = 9 - y^2 \implies x = \sqrt{9-y^2}$. For $x$ to be real, $9-y^2 \ge 0 \implies y^2 \le 9 \implies -3 \le y \le 3$. Since $y \ge 0$, we intersect these conditions to get $0 \le y \le 3$. Final answer: Domain = $[-3, 3]$, Range = $[0, 3]$.
• Q: Let $f = \{(1,1), (2,3), (0,-1), (-1,-3)\}$ be a linear function from $\mathbb{Z}$ to $\mathbb{Z}$. Find $f(x)$. A: Step 1: Let the linear function be $f(x) = mx + c$, where $m$ and $c$ are integers. Step 2: Use the point $(0,-1)$ to find $c$: $f(0) = m(0) + c = -1 \implies c = -1$. Step 3: Use another point, say $(1,1)$, to solve for $m$: $f(1) = m(1) - 1 = 1 \implies m = 2$. Step 4: Verify with a third point $(-1,-3)$: $f(-1) = 2(-1) - 1 = -3$, which is consistent. Final answer: The linear function is $f(x) = 2x - 1$.
• Q: Find the domain of the function $f(x) = \sqrt{x-1} + \frac{1}{\sqrt{7-x}}$. A: Step 1: Set up the constraint for the first term: $x-1 \ge 0 \implies x \ge 1$. This yields interval $D_1 = [1, \infty)$. Step 2: Set up the constraint for the second term. Since it is in the denominator, $7-x > 0 \implies x < 7$. This yields interval $D_2 = (-\infty, 7)$. Step 3: Find the intersection of both domains: $D = D_1 \cap D_2 = [1, \infty) \cap (-\infty, 7)$. Final answer: The domain is $[1, 7)$.

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