Problem 31 - A Complete Resource Book in Mathematics for JEE Main

Question

Number of integral solutions of \(\frac{x+2}{x^{2}+1}>\frac{1}{2}\) is (A) 0 (B) 1 (C) 2 (D) 3

Step-by-Step Solution

Answer

The number of integral solutions is 3 (Option D).

Step 1: Express Inequality in Terms of x

Start by manipulating the given inequality \( \frac{x+2}{x^{2}+1} > \frac{1}{2} \). Cross-multiply to eliminate the fractions: \( 2(x+2) > x^2 + 1 \). This simplifies to the expression \( 2x + 4 > x^2 + 1 \).

Step 2: Rearrange Inequality to Form a Quadratic Expression

Rearrange the expression to form a standard quadratic inequality: \( x^2 - 2x - 3 < 0 \). This expression needs to be solved to determine the interval of x satisfying the inequality.

Step 3: Factorize the Quadratic Expression

Factorize the quadratic expression \( x^2 - 2x - 3 \). This can be broken down into \( (x-3)(x+1) \). Therefore, the inequality \( (x-3)(x+1) < 0 \) needs to be solved.

Step 4: Determine Solution Intervals

Find critical points from the factors, which are \( x = 3 \) and \( x = -1 \). The expression \( (x-3)(x+1) < 0 \) holds true between these critical points.

Step 5: Check Integer Values Within the Interval

The interval where the inequality holds is \( -1 < x < 3 \). The integral values in this interval are \( 0, 1, \) and \( 2 \).

Step 6: Verify Integer Solutions

Verify each integer value by substituting them back into the original inequality. For \( x = 0, 1, \) and \( 2 \), the inequality holds as each satisfies \( \frac{x+2}{x^2+1} > \frac{1}{2} \). Thus, there are 3 integral solutions.

Key Concepts

Understanding Inequalities

Inequalities are mathematical expressions that describe the relative size or order of two values. In our original exercise, we dealt with an inequality of rational expressions: \( \frac{x+2}{x^2+1} > \frac{1}{2} \). To solve an inequality, the aim is to find a range of values for a variable that makes the inequality true. A helpful way to approach such problems is to first eliminate any fractions by cross-multiplying, as shown in the solution, leading to simpler terms you can work with.

Inequalities can sometimes be tricky because they do not behave like equalities. For example, when multiplying or dividing by a negative number, the inequality sign flips. Always keep a lookout for this rule as it often becomes crucial in solving these problems. In the context of our exercise, once cross-multiplication helps remove fractions, we're left with a quadratic inequality \( 2x + 4 > x^2 + 1 \), which we can rearrange into a standard quadratic form.

Exploring Quadratic Expressions

A quadratic expression is a type of polynomial that involves only terms up to the second degree, typically written as \( ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants. In our exercise, after rearranging, we obtained \( x^2 - 2x - 3 \). Each term in this expression tells us something about the shape and nature of the corresponding quadratic function.

• The \( x^2 \) term indicates it's a parabola.
• The linear term \(-2x\) impacts the location and direction of the parabola.
• The constant \(-3\) shifts the parabola up or down on the y-axis.

Quadratic inequalities require solving much like quadratic equations, except that instead of solving for \( x \) values that equal a specific number, we find ranges of \( x \) values that satisfy the inequality (either greater than or less than). This often involves factoring the quadratic, determining roots, and testing intervals between and beyond these roots.

Factoring Quadratics

Factoring is the process of breaking down an algebraic expression into a product of simpler terms, known as factors. For quadratics, this means writing the expression \( ax^2 + bx + c \) as \((x - p)(x - q)\) where \(p\) and \(q\) are the roots of the expression. In our problem, the expression \( x^2 - 2x - 3 \) was factored into \( (x-3)(x+1) \).

To factor quadratics effectively:

• Look for two numbers that multiply to \( c \) (the constant term) and add to \( b \) (the linear coefficient).
• In our case, the numbers \( -3 \) and \( 1 \) work because \( (-3)(1) = -3 \) and \( -3 + 1 = -2 \).

Once factored, solving the inequality \((x-3)(x+1) < 0\) can be done by identifying critical points \( x = 3 \) and \( x = -1 \). Testing intervals helps determine where the expression is negative. For this exercise, this means checking the sign of the product across different intervals along the x-axis, leading to the solution that only values between \( -1 \) and \( 3 \) satisfy the condition.