Problem 17 - Calculus for JEE ADVANCED

Question

$$ \begin{aligned} &f(x)=\frac{\sin ^{-1}(3-x)}{\log (|x|-2)} \\ &\text { Let } g(x)=\sin ^{-1}(3-x) \\ &\text { or }-1 \leq 3-x \leq 1 \\ &\text { The domain of } g(x) \text { is }[2,4] \text {. } \\ &\text { Let } h(x)=\log (|x|-2) \\ &\text { i.e., }|x|-2>0 \text { or }|x|>2 \\ &\text { i.e., } x<-2 \text { or } x>2 \\ &\therefore \quad \text { Domain }(-\infty,-2) \cup(2, \infty) \\ &\text { We know that } \\ &\qquad(f / g)(x)=\frac{f(x)}{g(x)} \forall x \in D_{1} \cap D_{2}-\\{x \in R: g(x)=0\\} \end{aligned} $$ Therefore, the domain of \(f(x)\) is \((2,4]-\\{3\\}=(2,3) \cup(3,4]\)

Step-by-Step Solution

Answer

The domain of \(f(x)\) is \((2,3) \cup (3,4]\).

Step 1: Identify Domain Restrictions for g(x)

Function \( g(x) = \sin^{-1}(3-x) \) has an input restriction based on the range of the arcsine function. Thus, \(-1 \leq 3-x \leq 1\). Solving for \(x\), we get \(2 \leq x \leq 4\). Hence, the domain of \( g(x) \) is \([2,4]\).

Step 2: Identify Domain Restrictions for h(x)

Function \( h(x) = \log(|x|-2) \) requires \(|x|-2 > 0\). Solving gives \(|x| > 2\), which means \(x < -2\) or \(x > 2\). Therefore, the domain of \( h(x) \) is \((-fty, -2) \cup (2, fty)\).

Step 3: Intersection for the Combined Domain

The function \( f(x) = \frac{g(x)}{h(x)} \) is defined where both \(g(x)\) and \(h(x)\) are defined and \(h(x) eq 0\). Thus, find the intersection \([2,4] \cap (2, fty) = (2,4]\).

Step 4: Exclude Points where Denominator is Zero

\(h(x) = \log(|x|-2)\) is zero when \(|x|-2 = 1\), giving \(|x| = 3\), thus \(x = 3\) as a solution. Therefore, exclude \(x = 3\) from the domain \((2,4]\).

Step 5: Finalize the Domain of f(x)

Combine the restrictions to find the domain of \( f(x) \) by excluding \(3\) from \((2,4]\). The final domain is \((2,3) \cup (3,4]\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles based on given trigonometric values. The function \( \sin^{-1}(x) \) is called arcsine and its domain is from \(-1\) to \(1\). This means that only values within this range are valid inputs for the arcsine function.
In our exercise, we have \( g(x) = \sin^{-1}(3-x) \). For this to be valid, \((3-x)\) must also lie between \(-1\) and \(1\). Solving the inequality \(-1 \leq 3-x \leq 1\) gives us that the domain for \(g(x)\) must be \([2,4]\).
Understanding the range restrictions of inverse trigonometric functions is crucial since it determines where the function is valid. Knowing the limitations can help avoid errors when working with them.

Logarithmic Functions

Logarithmic functions are defined as the inverse of exponential functions, commonly used in many fields to deal with exponential growth or decay. The function \( \log(x) \) requires \(x > 0\) because the logarithm of a non-positive number is undefined.
In our scenario, we have \( h(x) = \log(|x|-2) \). This formulation requires \(|x|-2 > 0\), so \(|x| > 2\). Therefore, this means \(x < -2\) or \(x > 2\). The domain for \(h(x)\) thus becomes \((-\infty, -2) \cup (2, \infty)\).
It's important to note that logarithmic functions are only defined with positive arguments, and special care must be taken to maintain such conditions in complex expressions.

Domain Restrictions

Domain restrictions dictate where functions are defined and valid. They are critical to understand the limits within which a function operates without producing undefined or non-valid outputs.
When working with functions like \(f(x) = \frac{\sin^{-1}(3-x)}{\log(|x|-2)}\), it's crucial to combine the separate domain restrictions from both components. \(g(x)\) and \(h(x)\) each have their specific domains: \([2,4]\) and \((-\infty, -2) \cup (2, \infty)\), respectively.
The overall domain of the function \(f(x)\) is the intersection of these domains, as the function is only defined where both components are valid. However, any values making the denominator zero must be excluded to avoid undefined conditions. This comprehensive approach ensures the function behaves as expected and is crucial for accurate problem-solving in mathematics.

Function Intersections

Function intersections help determine where multiple conditions or functions overlap, giving insight into valid domains or shared characteristics.
In our problem, we determine the domain of \(f(x)\) by finding the intersection of the domains of \(g(x)\) and \(h(x)\), resulting in \((2,4]\).
However, intersections may also demand excluding specific values where conditions cause issues. In this case, we exclude the point \(x = 3\), where the denominator equals zero, by identifying \(|x| = 3\) as a problematic value.
Identifying such intersections, inclusive and exclusive, can help solve complex functions with multiple expressions, providing a clearer understanding and accurate solutions.