Problem 154 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Domain is \((-1, 0) \cup (1, 2) \cup (2, \infty)\).

Step 1: Identify Valid Conditions for the Fraction

For the function \( f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x) \) to be defined, the denominator of the fraction \( 4-x^2 \) must not be equal to zero. Thus, we solve the equation \(4-x^2 = 0\). This gives \(x^2 = 4\), so \(x = \pm 2\). Therefore, \(x eq 2\) and \(x eq -2\).

Step 2: Identify Valid Conditions for the Logarithm

Next, consider the domain of the logarithmic function \( \log_{10}(x^3-x) \). The argument of the log, \(x^3-x\), must be positive, i.e., \(x^3-x > 0\). Solve the inequality by factoring: \(x(x^2-1) = x(x-1)(x+1) > 0\). The critical points are \(x = 0, 1, -1\).

Step 3: Test Intervals Determined by Critical Points

Test the sign of \(x(x-1)(x+1)\) in the intervals determined by the critical points: - For \((-\infty, -1)\), choose a test point like \(-2\), yielding negative result.- For \((-1, 0)\), choose a test point like \(-0.5\), yielding positive result.- For \((0, 1)\), choose a test point like \(0.5\), yielding negative result.- For \((1, \infty)\), choose a test point like \(2\), yielding positive result.

Step 4: Combine Results for Domain

The valid intervals for the logarithmic function are \((-1, 0)\) and \((1, \infty)\). Combined with the need for \(x eq \pm 2\), the domain of the function \(f(x)\) is \((-1, 0) \cup (1, 2) \cup (2, \infty)\).

Key Concepts

Domain of Definition

When addressing any mathematical function, determining its domain of definition is crucial. The domain of definition is the set of all possible inputs (or x-values) for which the function is defined and produces real numbers. In simpler terms, it's where the function 'works' without any hiccups.

For the function \( f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x) \), there are specific conditions we need to consider for both the fractional and logarithmic parts. It's important to note that:

• The denominator of any fraction must not be zero, as division by zero is undefined.
• The argument of a logarithm must always be positive because you can't take the logarithm of zero or a negative number within real numbers.

This means, we spend time identifying these conditions to determine where exactly the function can safely operate without landing in undefined territory.

Logarithmic Functions

Logarithmic functions, like \( \log_{10}(x^3-x) \), require careful consideration of their domain. The primary rule is that the argument (in this case, \( x^3-x \)) must be positive. This arises because logarithms of zero or negative numbers are undefined within real numbers.

To find the domain for \( \log_{10}(x^3-x) \), we solve the inequality \( x^3-x > 0 \). This involves:

• Factoring the expression as \( x(x^2-1) = x(x-1)(x+1) \).
• Determining the critical points: \( x = 0, 1, -1 \).
• Testing intervals between these points to ensure positivity.

By testing these intervals, we confirm where the expression is positive and thus where the logarithm is defined. This results in valid intervals such as \((-1, 0)\) and \((1, \infty)\). Removing points where the expression turns negative, ensures our function remains within its defined boundaries.

Rational Functions

Rational functions, particularly the function \( \frac{3}{4-x^2} \), necessitate the denominator never being zero. If it were zero, the function would be undefined, due to the impossibility of division by zero.

To address this, we:

• Set the denominator \( 4-x^2 eq 0 \).
• Solving \( 4-x^2 = 0 \), we find the critical points \( x = \pm 2 \).

These points are excluded from the domain, as substituting them into the function would result in division by zero. Thus, ensuring \( x eq 2 \) and \( x eq -2 \), dictates our function is safe to evaluate for all other x-values, aligning with other constraints from the logarithmic part for the combined domain.

Inequalities in Intervals

To effectively understand and manipulate the function's domain, we must work with inequalities within intervals. In this exercise, testing intervals between critical points allows us to determine where the expression \( x(x-1)(x+1) \) remains positive, thus acceptable for our logarithmic function.

We take the following approach:

• Identify critical points: \( x = -1, 0, 1 \).
• Test signs in intervals created by these points, such as \((-\infty, -1)\), \((-1, 0)\), \((0, 1)\), and \((1, \infty)\).
• Determine positivity in intervals yields valid intervals for the domain.

This step helps exclude regions where conditions for the rational or logarithmic part fail, guiding us to establish the function's complete domain. Here, recognizing both opportunities and constraints ensures we don't fall into undefined value traps.