JEE Advanced 2001

LEVELJEE Main

Animated Solution for Mathematics - Functions: The domain of definition of $f (x) = \frac{l o g _{2} ( x + 3 )}{x ^{2} + 3 x + 2}$ is

Q: Why is it $x > -3$ and not $x \geq -3$?

Remember, the logarithm function $\log_b(y)$ is only defined for $y > 0$. If $x = -3$, the argument becomes $0$, and $\log_2(0)$ is undefined (it approaches $-\infty$). We must keep the argument strictly positive.

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Visualized Solution

Introduction to Domain Analysis

To find the domain of $f (x) = \frac{l o g _{2} ( x + 3 )}{x ^{2} + 3 x + 2}$ , we need to find all valid values of $x$ .
We must check two main conditions:
1. The logarithm in the numerator must be defined.
2. The denominator must not be zero.

Logarithm Constraint: $lo g_{2} (x + 3)$

Let's analyze the numerator: $lo g_{2} (x + 3)$ .
Rule: The argument of any logarithm must be strictly positive.

Setting up the Log Inequality

Applying the rule to our function:
$x + 3 > 0$

Solving the Log Inequality

Solving for $x$ :
$x > - 3$
This means $x$ must lie in the interval $(- 3, \infty)$ .

Denominator Constraint: $x^{2} + 3 x + 2 \neq = 0$

Now, let's look at the denominator: $x^{2} + 3 x + 2$ .
Rule: Division by zero is undefined in mathematics.

Setting up the Denominator Condition

To prevent division by zero, we enforce:
$x^{2} + 3 x + 2 \neq = 0$

Factorizing the Quadratic

We need to factorize $x^{2} + 3 x + 2$ .
Splitting the middle term:
$x^{2} + 2 x + x + 2 \neq = 0$

Completing the Factorization

Grouping the terms:
$x (x + 2) + 1 (x + 2) \neq = 0$
$(x + 1) (x + 2) \neq = 0$

Identifying Forbidden Points

For the product to be non-zero, neither factor can be zero:
$x + 1 \neq = 0 \Rightarrow x \neq = - 1$
$x + 2 \neq = 0 \Rightarrow x \neq = - 2$

Visualizing the Forbidden Points

We must remove $x = - 1$ and $x = - 2$ from our valid region.
Let's mark these as "holes" on our number line.

Final Domain Intersection

Combining all constraints:
Condition 1: $x \in (- 3, \infty)$
Condition 2: $x \in / {- 1, - 2}$
Final Domain: $(- 3, \infty) ∖ {- 1, - 2}$

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The Sigma Insight: Domain and Range of a Function

Solution Diagram

Analyzing the Setup

To find the domain of the function

f (x) = \frac{lo g _{2} ( x + 3 )}{x ^{2} + 3 x + 2}

we must identify all values of

x

for which the expression is well-defined. This requires satisfying two primary mathematical constraints simultaneously.

The Logarithm's Secret

The numerator contains a logarithmic term,

lo g_{2} (x + 3)

. Logarithms are only defined for strictly positive arguments.

Therefore, we must satisfy the inequality:

x + 3 > 0

Solving for

x

, we find:

x > - 3

This establishes our primary interval of existence as

(- 3, \infty)

The Denominator's Trap

The denominator,

x^{2} + 3 x + 2

, cannot be equal to zero, as division by zero is undefined. We must identify the roots of this quadratic expression to exclude them from our domain.

Factoring the quadratic, we get:

x^{2} + 3 x + 2 = (x + 1) (x + 2)

Setting the factors to zero, we find the forbidden values:

(x + 1) (x + 2) = 0 \Rightarrow x = - 1, x = - 2

Both

x = - 1

and

x = - 2

fall within the interval

(- 3, \infty)

established in the previous step. Consequently, these points must be excluded.

Final Calculation

To determine the final domain, we take the intersection of the interval

x > - 3

and the exclusion set $x eq -1, -2$.

We start with the interval

(- 3, \infty)

and remove the points

{- 2, - 1}

The resulting domain is expressed as:

$(- 3, - 2) \cup (- 2, - 1) \cup (- 1, \infty)$

Alternatively, this can be written in set notation as:

$(- 3, \infty) ∖ {- 2, - 1}$

Common Traps & FAQs

Why can't we just simplify the expression first? If I factor the denominator, can't I cancel things out?

Great question! In domain analysis, you must look at the function in its *original* form. If you cancel terms, you might accidentally 'hide' a forbidden point where the function was originally undefined. Always analyze the domain *before* simplifying!

Why is it

x > - 3

and not

x \geq - 3

Remember, the logarithm function

lo g_{b} (y)

is only defined for

y > 0

. If

x = - 3

, the argument becomes

0

, and

lo g_{2} (0)

is undefined (it approaches

- \infty

). We must keep the argument strictly positive.

Why do we exclude

- 1

and

- 2

even if they are greater than

- 3

The domain is the set of *all* values that satisfy *all* conditions simultaneously. Even if a value is greater than

- 3

, if it makes the denominator zero, it breaks the function. We must exclude any value that violates *any* of our rules.

Animated Solution for Mathematics - Functions: The domain of definition of $f (x) = \frac{l o g _{2} ( x + 3 )}{x ^{2} + 3 x + 2}$ is

Select Answer:

Visualized Solution

Introduction to Domain Analysis

Logarithm Constraint: $lo g_{2} (x + 3)$

Setting up the Log Inequality

Solving the Log Inequality

Denominator Constraint: $x^{2} + 3 x + 2 \neq = 0$

Setting up the Denominator Condition

Factorizing the Quadratic

Completing the Factorization

Identifying Forbidden Points

Visualizing the Forbidden Points

Final Domain Intersection

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Logarithm's Secret

The Denominator's Trap

Final Calculation

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of definition of f(x)=x2+3x+2log2​(x+3)​ is

Select Answer:

Visualized Solution

Introduction to Domain Analysis

Logarithm Constraint: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } log2​(x+3)

Setting up the Log Inequality

Solving the Log Inequality

Denominator Constraint: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+3x+2=0

Setting up the Denominator Condition

Factorizing the Quadratic

Completing the Factorization

Identifying Forbidden Points

Visualizing the Forbidden Points

Final Domain Intersection

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Logarithm's Secret

The Denominator's Trap

Final Calculation

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The domain of the definition of the function $f (x) = \frac{1}{4 - x ^{2}} + lo g_{10} (x^{3} - x)$ is :-

The domain of the function $f (x) = \frac{c o s ^{- 1} ( \frac{x ^{2} - 5 x + 6}{x ^{2} - 9} )}{l o g _{e} ( x ^{2} - 3 x + 2 )}$ is

The domain of definition of the function $y = \frac{1}{l o g _{10} ( 1 - x )} + x + 2$ is

The domain of the function $f (x) = \frac{s i n ^{- 1} ( x - 3 )}{9 - x ^{2}}$ is

Let $f (x) = lo g_{e} x$ and $g (x) = \frac{x ^{4} - 2 x ^{3} + 3 x ^{2} - 2 x + 2}{2 x ^{2} - 2 x + 1}$ . Then the domain of $f (g (x))$ is

The domain of definition of the function given by the equation $2^{x} + 2^{y} = 2$ is

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The domain of the function $f (x) = \frac{1}{[ x ] ^{2} - 3 [ x ] - 10}$ is (where $[x]$ denotes the greatest integer less than or equal to $x$ )