Problem 158 - A Complete Resource Book in Mathematics for JEE Main

Question

The domain of the function \(f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^{2}}}\) is (A) \([2,3]\) (B) \([2,3)\) (C) \([1,2]\) (D) \([1,2)\)

Step-by-Step Solution

Answer

(A) \([2, 3]\)

Step 1: Analyze the Domain of the Inverse Sine Function

The function involves \(\sin^{-1}(x-3)\). The domain of \(\sin^{-1}(y)\) is \([ -1, 1 ]\). Therefore, for \(\sin^{-1}(x-3)\) to be defined, \[ -1 \leq x - 3 \leq 1 \] solve for \(x\):1. Adding 3 gives: \[ 2 \leq x \leq 4 \] describes the range of \(x\) for which \(\sin^{-1}(x-3)\) is valid.

Step 2: Analyze the Domain of the Square Root Function in Denominator

The denominator contains \(\sqrt{9-x^2}\). The square root is defined when its argument is non-negative: \[ 9-x^2 \geq 0 \]Solving gives:1. Rearrange to \( x^2 \leq 9 \).2. Taking the square root on both sides, we get: \[ -3 \leq x \leq 3 \].

Step 3: Find the Intersection of Both Domains

To ensure both functions \(\sin^{-1}(x-3)\) and \(\sqrt{9-x^2}\) are valid, their domains must intersect. From Step 1, the domain is \([2, 4]\) and from Step 2, it's \([-3, 3]\). The overlap (intersection) of these is:\[ x \in [2, 3] \] indicating where both conditions are true simultaneously.

Step 4: Conclusion

The function \(f(x)=\frac{\sin^{-1}(x-3)}{\sqrt{9-x^{2}}}\) is defined on the intersection of the domains found above. Therefore, the overal domain for \(f(x)\) is the intersection: \([2, 3]\). This corresponds to option (A).

Key Concepts

Inverse Sine Function

The inverse sine function, denoted as \(\sin^{-1}(x)\), or arcsine, is crucial when dealing with angles. Unlike regular sine, which provides the ratio of the opposite side to the hypotenuse in a right-angled triangle, arcsine takes a ratio and returns an angle. For it to provide real numbers, the input must lie within the range

• -1
• to 1

This is essential for understanding its domain. For instance, in functions where you see \(\sin^{-1}(x-3)\), it requires that \(-1 \leq x-3 \leq 1\). Simplifying this gives:

• \(2 \leq x \leq 4\)

This range ensures that each input allows for a real output from the arcsine function.

Square Root Function

Square root functions involve sqrt(), like \(\sqrt{9-x^2}\) in our case. For such a function to stay within the realm of real numbers, the expression inside must be non-negative. This translates to solving inequalities like \(9 - x^2 \geq 0\). Here's a simpler way to visualize:

• \(x^2 \leq 9\)

The solution, \(-3 \leq x \leq 3\), indicates we're dealing with a range of values for \(x\). Any \(x\) within this span will keep the square root valid. This becomes crucial when intersecting with other domain requirements.

Domain Intersection

Now that we know both functions' domains, find where they overlap. This is called the domain intersection. For our situation:

• From the inverse sine segment, \(x\) ranges from 2 to 4.
• The square root permits from -3 to 3.

The overlap comes out cleanly as \(2 \leq x \leq 3\). Join these together for a clear intersection: only values from 2 to 3 work in both parts, making \([2, 3]\) the function's domain. Recognizing these common values ensures both parts of the fraction are valid, letting the whole expression be defined and solvable.