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Q3:
Find the domain of the function $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

Solution :

Step 1: Theoretical Foundation of Domain

The domain of a rational function $f(x) = \frac{P(x)}{Q(x)}$ consists of all real numbers $x \in \mathbb{R}$ such that the denominator $Q(x) \neq 0$. Any value of $x$ that results in $Q(x) = 0$ is excluded from the domain to avoid division by zero, which is undefined in the set of real numbers [Per the definition of functions in $\mathbb{R}$].

Step 2: Identifying the Denominator

Given $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$, the denominator is defined as the quadratic polynomial:

$Q(x) = x^2 - 8x + 12$

Step 3: Solving for Roots of the Denominator

To find the excluded values, we set $Q(x) = 0$ and solve for $x$ using the factorization method [By the Zero Product Property]:

$x^2 - 8x + 12 = 0$

We seek two numbers that multiply to $12$ and add to $-8$. These numbers are $-6$ and $-2$.

$x^2 - 6x - 2x + 12 = 0$

$x(x - 6) - 2(x - 6) = 0$

$(x - 2)(x - 6) = 0$

Therefore, $x - 2 = 0 \implies x = 2$ or $x - 6 = 0 \implies x = 6$.

Step 4: Establishing the Domain Interval

The function $f(x)$ is defined for all real numbers except for the set of points where the denominator vanishes. Thus, the domain is the set of all real numbers excluding $\{2, 6\}$.

2 6 Domain: All real numbers except x=2 and x=6

Step 5: Final Set-Builder Notation

Expressing the result in standard set notation:

$Domain(f) = \mathbb{R} \setminus \{2, 6\}$

Alternatively, in interval notation:

$Domain(f) = (-\infty, 2) \cup (2, 6) \cup (6, \infty)$

Final Solution: The domain of the function is the set of all real numbers except $2$ and $6$, represented as $\mathbb{R} - \{2, 6\}$.


More Questions from Class 11 Mathematics Relations and Functions Miscellaneous Exercise on Chapter 2


CBSE Solutions for Class 11 Mathematics Relations and Functions


Chapters in CBSE - Class 11 Mathematics


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