Problem 43 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function $$ f(x)=\sin ^{-1}\left(x-x^{2}\right)+\sqrt{1-\frac{1}{|x|}}+\frac{1}{\left[x^{2}-1\right]} $$ is defined in the interval (where \([\cdot]\) is the greatest integer) (A) \(x \in\left(\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\) (B) \(x \in\left(1, \frac{1+\sqrt{5}}{2}\right)\) (C) \(x \in\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]\) (D) \(x \in\left(-\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\)

Step-by-Step Solution

Answer

The function is defined in the interval (B) \( x \in \left(1, \frac{1+\sqrt{5}}{2}\right) \).

Step 1: Analyzing the Inverse Sine Function

The inverse sine function, denoted as \( \sin^{-1}(y) \), requires that its argument \( y = x - x^2 \) must be between -1 and 1, inclusive. This gives the inequality \(-1 \leq x - x^2 \leq 1\). We'll need to solve this inequality to determine the possible values for \( x \).

Step 2: Solving Inequality for Inverse Sine

First, set up the inequality for the inverse sine argument: \( x - x^2 \geq -1 \). Simplifying, we find \( x^2 - x - 1 \leq 0 \). Solving this quadratic inequality, we find the solutions are approximately \( x \in \left[\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right] \). For the other side \( x - x^2 \leq 1 \), rearranging gives \( x^2 - x + 1 \geq 0 \), which is always true for complex roots, so it adds no restriction.

Step 3: Analyzing the Square Root Function

For \( \sqrt{1 - \frac{1}{|x|}} \) to be real, the expression inside the square root \( 1 - \frac{1}{|x|} \) must be non-negative. Thus, \( |x| \geq 1 \). Since we are interested in positive values of \( x \), this becomes \( x \geq 1 \).

Step 4: Identifying Domain Restrictions from the Greatest Integer Function

The expression \( \frac{1}{[x^2 - 1]} \) requires that \( x^2 - 1 \) must not be an integer 0, so \( x^2 eq 1 \). This implies that \( x eq \pm 1 \). However, \( x \geq 1 \) from our earlier analysis narrows this to \( x > 1 \).

Step 5: Intersection of Domain Conditions

Combining all conditions: 1. From inverse sine, \( x \in \left[\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right] \). 2. From the square root function, \( x \geq 1 \). 3. From the greatest integer function, \( x eq 1 \). The intersection of these is \( x \in \left(1, \frac{1 + \sqrt{5}}{2}\right)\).

Step 6: Finalizing the Interval

The appropriate interval that satisfies all domain requirements for \( f(x) \) is given in choice (B): \( x \in \left(1, \frac{1 + \sqrt{5}}{2}\right) \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, such as the inverse sine function, refer to the inverse operations of standard trigonometric functions. These functions take a real number from a specific range and return an angle. For instance, the inverse sine function, denoted as \( \sin^{-1}(x) \), can only take input values between \(-1\) and \(1\), inclusive. This range restriction stems from the fact that the sine of an angle can only produce values within these bounds.

When given a function with an inverse trigonometric component, like \( f(x) = \sin^{-1}(x - x^2) \), we must ensure that the expression \( x - x^2 \) stays within the \([-1, 1]\) range. To find these valid \(x\) values, we solve the inequality \(-1 \leq x - x^2 \leq 1\). Through quadratic inequality solving, determining this range allows us to understand the permissible values for \(x\) in relation to the inverse function's constraints.

Quadratic Inequalities

Quadratic inequalities are expressions where a quadratic polynomial is set in relation to zero with symbols like \(<\), \(\leq\), \(>\), or \(\geq\). Solving these inequalities often involves finding the roots of the corresponding quadratic equation and testing intervals determined by these roots.

For the expression derived from the inverse sine requirement, \(x^2 - x - 1 \leq 0\), we identify the roots using the quadratic formula, giving us solutions \( x = \frac{1 \pm \sqrt{5}}{2} \). The valid \(x\) values are those which satisfy the inequality within these roots. Interval testing confirms that the solution includes \( x \in \left[\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right] \). This method is frequently used when dealing with inverse trigonometric functions or any other scenarios where a range of validity must be established.

Square Root Restrictions

Expressions involving square roots require careful handling to ensure the value inside the square root, known as the radicand, is non-negative. If negative, the square root is undefined in real numbers because of the principal restriction of real roots to non-negative values.

The function \( \sqrt{1 - \frac{1}{|x|}} \) indicates the expression \(1 - \frac{1}{|x|} \) must be non-negative, thus \( |x| \geq 1 \). Since the focused domain considers positive \(x\), this simplifies to \( x \geq 1 \). This condition intersecting with other requirements sharpens our understanding of the domain choices for \(f(x)\), restricting values further to ensure we only work with valid real numbers.

Greatest Integer Function

The greatest integer function, often symbolized by \([ \cdot ]\), refers to the mathematical operation that takes any real number and outputs the largest integer less than or equal to that number. For example, \([3.7] = 3\), while \([-1.2] = -2\). This function introduces additional considerations within domain restrictions.

When dealing with expressions involving the greatest integer function, we must be cautious of points where it might create division by zero or undefined behavior. In the exercise, \( \frac{1}{[x^2 - 1]} \) suggests such a condition. Specifically, \(x^2 - 1\) must never equal zero as it would result in an undefined expression. This translates into \(x \, eq \, \pm 1\). However, since we've already determined \(x > 1\), only \(x eq 1\) remains relevant, ensuring proper domain determination for the provided function.