Problem 154 - A Complete Resource Book in Mathematics for JEE Main

Question

Domain of definition of the function \(f(x)=\frac{3}{4-x^{2}}+\log _{10}\left(x^{3}-x\right)\), is (A) \((1,2)\) (B) \((-1,0) \cup(1,2)\) (C) \((1,2) \cup(2, \infty)\) (D) \((-1,0) \cup(1,2) \cup(2, \infty)\)

Step-by-Step Solution

Answer

The domain is (C) \((1, 2) \cup (2, \infty)\).

Step 1: Identify Restrictions for Denominator

The function \( f(x) \) contains a term \( \frac{3}{4-x^2} \). This denominator must not be zero. Solve \( 4-x^2 = 0 \) to find where the function is undefined: \( x^2 = 4 \). Thus, \( x = 2 \) or \( x = -2 \). So, \( x eq 2 \) and \( x eq -2 \). These values must be excluded from the domain.

Step 2: Identify Restrictions for Logarithm

The function contains \( \log_{10}(x^3 - x) \). The argument of a logarithm, \( x^3 - x \), must be positive. Solve \( x^3 - x > 0 \).Factor the inequality as \( x(x-1)(x+1) > 0 \). Determine the sign changes by testing intervals between the roots: \(-1, 0, \) and \(1\). - For \( x < -1 \), choose \( x = -2 \): \( (-2)((-2)-1)((-2)+1) = (-)(-)(-) = - \) (negative).- For \(-1 < x < 0\), choose \( x = -0.5 \): \( (-0.5)((-0.5)-1)((-0.5)+1) = (-)(-)(-) = - \) (negative).- For \( 0 < x < 1 \), choose \( x = 0.5 \): \( (0.5)((0.5)-1)((0.5)+1) = (+)(-)(+) = - \) (negative).- For \( x > 1 \), choose \( x = 2 \): \( (2)(2-1)(2+1) = (+)(+)(+) = + \) (positive).Thus, \( x > 1 \).

Step 3: Combine Restrictions

Consider the conditions from both steps: 1. \( x eq 2 \) and \( x eq -2 \) (from the denominator restriction).2. \( x > 1 \) (from the logarithm restriction).Therefore, the domain of \( f(x) \) is \( (1, 2) \cup (2, \infty) \).

Step 4: Match with Given Options

Compare the derived domain \( (1, 2) \cup (2, \infty) \) with the answer options. The correct match is option (C).

Key Concepts

Functions

Understanding functions is vital as they are a fundamental concept in mathematics that describe relationships between variables. Think of a function as a machine that takes an input, processes it, and gives a unique output. For each input value, there is one and only one output value.
One example we deal with here is the function \( f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x) \). This function comprises different parts: a rational function and a logarithmic function, each imposing their own constraints on the possible input (or **domain**).

• The rational part \( \frac{3}{4-x^2} \) cannot have a denominator of zero, hence we need to exclude values that make \( 4-x^2 = 0 \).
• The logarithmic part \( \log_{10}(x^3-x) \) requires the argument \( x^3-x \) to be positive because the logarithm of non-positive numbers is undefined.

This is why dissecting the components of functions and understanding their requirements is crucial in determining the domain.

Logarithms

Logarithms are mathematical operations that help solve equations where the unknown appears in the exponent. They essentially "undo" the process of exponentiation, allowing us to find the power to which a number must be raised to obtain another number. In simpler terms, if you have \( b^y = x \), then \( y = \log_b(x) \).
In the function \( f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x) \), the logarithmic component \( \log_{10}(x^3-x) \) requires that \( x^3-x > 0 \). This restriction is essential because the logarithm

• is only defined for positive arguments - you can't take the logarithm of zero or a negative number.
• Therefore, when analyzing the domain of a function involving a logarithm, we need to solve the inequality \( x^3-x > 0 \) to find valid input values.

Solving this inequality involves finding the roots and determining where the product is positive. As discovered, this occurs when \( x > 1 \). Understanding the behavior of logarithms is crucial in these assessments.

Inequalities

Inequalities are expressions that dictate the relationships between two values, showing how one value is greater or lesser than the other. Solving inequalities is crucial in determining the domain of functions.
In the exercise, the inequality \( x^3-x > 0 \), associated with the logarithmic portion of the function, is pivotal. To solve it, we factor the expression as \( x(x-1)(x+1) > 0 \) and analyze different intervals.

• First, identify the critical points or roots: \(-1, 0,\) and \(1 \).
• Then test the sign of the expression in the intervals created by these points.
• The intervals determine where the expression is positive or negative. Only positive intervals are part of the valid solutions.

By solving such inequalities, we derive that the permitted territory for the logarithmic function is \( x > 1 \). This procedure ensures we only consider values where the function is valid, thereby maintaining mathematical correctness.