Problem 64 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The domain of the function $$ f(x)=\sqrt{\frac{-\log _{03}(x-1)}{-x^{2}+3 x+18}} $$ (A) \([2,6]\) (B) \((2,6)\) (C) \([2,6)\) (D) None of these

Step-by-Step Solution

Answer

The correct option is (D) None of these.

Step 1: Analyze the Square Root

The function is defined only when the expression inside the square root is non-negative. Therefore, we need to ensure that \( \frac{-\log_{03}(x-1)}{-x^{2}+3x+18} \geq 0 \). This means both the numerator \(-\log_{03}(x-1)\) and denominator \(-x^{2}+3x+18\) need to comply with certain rules.

Step 2: Evaluate the Logarithm Condition

The logarithmic function \( \log_{03}(x-1) \) is defined when \(x-1 > 0\), meaning \(x > 1\). However, since the base of the logarithm is not valid \(0.3\) (a non-standard use, implying possibly \(\log_{10}\) behavior), we will approximate when \(-\log_{10}(x-1)\) is still meaningful: \(x > 1\). Generally, we consider a natural base assumption.

Step 3: Determine Positive Denominator

For the denominator to be negative and meaningful, we solve \(-x^2 + 3x + 18 < 0\) to keep it negative. This translates to finding when the quadratic \(x^2 - 3x - 18 < 0\) by finding its roots. Solving \(x^2 - 3x - 18 = 0\) using the quadratic formula gives roots around \(6\) and \(-3\), so the quadratic is negative between \(x = 3\) and \(x = 6\).

Step 4: Combine Interval Results

From the steps above, the solution for the domain is restricted by the condition \(x > 3\) from Step 2 and \(3 < x < 6\) from Step 3. These interpret to an overall intersection constraint where the domain is \((3, 6)\).

Step 5: Find Overlapping Domain with Choices

Comparing the derived domain \((3, 6)\) with provided options, only none option matches precisely. Given the complexities and assumptions observed, realistic overlaps are: \((3, 6)\) does not fit \((2, 6)\) or \([2, 6)\) especially if considering an exclusive logarithm domain interpretation.

Key Concepts

Logarithmic Conditions

Understanding logarithmic functions is essential to determine function domains effectively. Here, we're dealing with the expression \(-\log_{03}(x-1)\).
This expression implies several things:

• Because \(\log(x-1)\) involves a logarithm, \(x-1\) must be greater than zero. Thus, \(x > 1\).
• The base 0.3 mentioned doesn't make sense, so it's safe to assume we're dealing with a regular log base, like 10.
• We need the overall expression in the fraction \(-\log_{10}(x-1)\) to validate the conditions for non-negativity.

Logarithmic conditions require the input (inside the parenthesis) to be positive for the function to be defined.
Remember, you cannot take the log of zero or a negative number.

Quadratic Inequalities

In this exercise, the quadratic inequality \(x^2 - 3x - 18 < 0\) plays a critical role. Quadratic inequalities help determine intervals of x-values where the expression holds true:

• To solve \(x^2 - 3x - 18 = 0\), we use the quadratic formula. The roots are around \(-3\) and \(+6\).
• The inequality is negative between its roots, resulting in the interval \(3 < x < 6\).
• This means, for the function's domain, x must be strictly between these values for the quadratic expression to be negative.

It's important to express solutions to quadratic inequalities in proper interval notation.
This helps in identifying the appropriate range of x-values that maintain the condition the inequality defines.

Interval Notation

To accurately express the valid domain of a function, interval notation is used. Here's why it matters:

• Interval notation simplifies the presentation of continuous ranges of numbers.
• Closed intervals [a, b] imply that both endpoints are included in the set.
• Open intervals (a, b) mean neither endpoint is included.
• Half-open intervals like [a, b) or (a, b] indicate inclusion of one endpoint.

In the given exercise, after applying both the logarithmic condition and quadratic inequality, the valid domain for x is \(3 < x < 6\).
This is written in interval notation as \(3, 6)\).
The importance of choosing correct endpoints based on the mathematical conditions ensures accuracy in determining solution sets for various functions.