Problem 26 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

The domain of the real valued function \(f(x)=\sqrt{1-2 x}+2 \sin ^{-1}\left(\frac{3 x-1}{2}\right)\) is \mathrm{\\{} \text {\\{} \text {\\{} K e r a l a ~ P E T - 2 0 0 7 ] ~ (a) \(\left[\frac{-1}{3}, 1\right]\) (b) \(\left[\frac{1}{2}, 1\right]\) (c) \(\left[\frac{-1}{2}, \frac{1}{3}\right]\) (d) \(\left[\frac{-1}{3}, \frac{1}{2}\right]\)

Step-by-Step Solution

Answer

The domain is \(\left[\frac{-1}{3}, \frac{1}{2}\right]\), so the answer is option (d).

Step 1: Analyze the Square Root Term

The function has a square root, so we need the expression under the square root to be non-negative for real values. This requires:\[1 - 2x \geq 0\]Solving gives:\[2x \leq 1\]\[x \leq \frac{1}{2}\]So, the domain from this condition is all \(x\) such that \( x \leq \frac{1}{2}\).

Step 2: Analyze the Inverse Sine Term

The \( \sin^{-1} \) function is defined for inputs in the range [-1, 1]. Thus,\[-1 \leq \frac{3x - 1}{2} \leq 1\]We split this compound inequality into two parts: 1. \[-1 \leq \frac{3x - 1}{2}\] gives \(3x - 1 \geq -2\) which simplifies to \(3x \geq -1\) and \(x \geq \frac{-1}{3}\).2. \[\frac{3x - 1}{2} \leq 1\] gives \(3x - 1 \leq 2\) which simplifies to \(3x \leq 3\) and \(x \leq 1\).

Step 3: Combine the Conditions

We need to combine the intervals obtained from both analyses:- From the square root term: \(x \leq \frac{1}{2}\)- From the inverse sine term: \(\frac{-1}{3} \leq x \leq 1\)The intersection of \([\frac{-1}{3}, 1]\) and \(( -\infty, \frac{1}{2}]\) is \(\left[\frac{-1}{3}, \frac{1}{2}\right]\).

Step 4: Choose the Correct Answer

From the previous step, we discovered the domain of \(f(x)\) is \(\left[\frac{-1}{3}, \frac{1}{2}\right]\). Therefore, the correct choice from the options given is \((d)\).

Key Concepts

Square Root Function

In understanding the domain of a function that includes a square root, it’s vital to recall that the expression under the square root must be non-negative. This is because you can only take the square root of zero or positive numbers without delving into complex numbers.
For the function given, the square root term is \( \sqrt{1 - 2x} \). For this to remain valid:

• The expression under the square root, \( 1 - 2x \), must be greater than or equal to zero.
• This results in the inequality: \( 1 - 2x \geq 0 \).
• Solving this inequality gives: \( x \leq \frac{1}{2} \).

Once you find this constraint, you've secured a critical part of determining the function's domain. Always ensure when dealing with square roots, to handle potential negative results cautiously by establishing this inequality.

Inverse Sine Function

The inverse sine function, known as \( \sin^{-1}(x) \), is another crucial element in determining the domain of a function. When dealing with this function, keep in mind:

• The input values for \( \sin^{-1}(x) \) must be within the range of \([-1, 1]\).
• For the function part \( \sin^{-1}\left(\frac{3x - 1}{2}\right) \), the condition becomes: \, \[-1 \leq \frac{3x - 1}{2} \leq 1\, \].

To solve this compound inequality, handle each part separately:

• \(-1 \leq \frac{3x - 1}{2} \) simplifies to \( x \geq \frac{-1}{3} \).
• \( \frac{3x - 1}{2} \leq 1 \) simplifies to \( x \leq 1 \).

Combining these results, you have the range \( \left[ \frac{-1}{3}, 1 \right] \), which reflects the possible x-values for the inverse sine component.

Compound Inequality

A compound inequality involves two statements joined by the word 'and' or 'or'. In our function, we combine the constraints obtained from both the square root and inverse sine function using 'and'.
To complete this process:

• From the square root condition, we know \( x \leq \frac{1}{2} \).
• From the inverse sine condition, we have \( \frac{-1}{3} \leq x \leq 1 \).

The key is to find the intersection of these conditions—and only those values that satisfy both inequalities simultaneously are valid in the function’s domain. The common interval from both is:

• \( \left[ \frac{-1}{3}, \frac{1}{2} \right] \)

This intersection gives the domain where the function is defined, ensuring no part of the conditions contradicts the other for valid real-number outputs.