Problem 13 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function \(f(x)=\frac{1}{u^{2}+u-2}\), where \(u=\frac{1}{x-1}\), is discontinuous at the points (A) \(x=-2,1, \frac{1}{2}\) (B) \(x=\frac{1}{2}, 1,2\) (C) \(x=1,0\) (D) None of these

Step-by-Step Solution

Answer

(B) \(x = \frac{1}{2}, 1, 2\).

Step 1: Substitute for u

Substitute \( u = \frac{1}{x-1} \) into \( f(x) = \frac{1}{u^2 + u - 2} \) to find a new function in terms of \( x \).

Step 2: Identify the points of discontinuity

The function will be discontinuous where the denominator \( u^2 + u - 2 = 0 \). Solve \( u^2 + u - 2 = 0 \) to find the values of \( u \) where the function is undefined.

Step 3: Solve the quadratic equation

Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -2 \) to solve \( u^2 + u - 2 = 0 \). The solutions are \( u = 1 \) and \( u = -2 \).

Step 4: Find corresponding x-values

Convert the values of \( u \) back to \( x \) using \( u = \frac{1}{x-1} \).For \( u = 1 \), solve \( 1 = \frac{1}{x-1} \) which gives \( x = 2 \).For \( u = -2 \), solve \( -2 = \frac{1}{x-1} \) which gives \( x = \frac{1}{2} \).

Step 5: Consider the value where x leads to undefined u

Recall that \( u = \frac{1}{x-1} \). The function is also undefined when \( x = 1 \) because division by zero occurs.

Step 6: Determine the points of discontinuity

Combine the possible x-values of discontinuity from the previous steps: \( x = 1, 2, \frac{1}{2} \).

Step 7: Select the correct option

Refer to the given options to match the discontinuous points \( x = 1, 2, \frac{1}{2} \). The correct answer is option (B) \( x = \frac{1}{2}, 1, 2 \).

Key Concepts

Quadratic Equation

A quadratic equation is a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are known constants, where \( a eq 0 \). Quadratic equations are fundamental in algebra and are used to find the roots, which are values of \( x \) that satisfy the equation.
To solve these equations, the quadratic formula is a reliable method: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula calculates the roots by substituting the coefficients \( a \), \( b \), and \( c \) from the equation into it.
Other methods to solve quadratic equations include:

• Factoring, where the quadratic is expressed as a product of binomials.
• Completing the square, which rebalances the equation to make one side a perfect square trinomial.

Quadratic equations play a key role in understanding functions that model real-world situations, such as projectile motion and area optimization.

Function Substitution

Function substitution involves replacing a variable in a function with another expression to simplify or solve the problem. This method is useful for transforming complex equations into more manageable forms.
For example, if you have \( f(x) = \frac{1}{u^2 + u - 2} \) where \( u = \frac{1}{x-1} \), you substitute \( u \) into the equation, altering the function's form to firmly relate it to \( x \).
Why use substitution?

• It helps in isolating variables, making it easier to solve equations.
• It can transform a given problem into a familiar structure, such as a quadratic.
• Substitution is often used in integration and differentiation in calculus.

This technique is quite powerful in breaking down obstacles in equations, providing a clearer picture of the relationship between variables.

Roots of Equations

Roots of an equation are the values of the variable that satisfy the equation, making it true. In other words, they are the x-values where the function crosses or touches the x-axis.
For a quadratic equation \( ax^2 + bx + c = 0 \), the roots can be found using various methods like:

• The quadratic formula, which is specifically used for quadratic equations.
• Graphical methods, where the intersection points of the graph with the x-axis are observed.
• Factoring and solving, which involves expressing the quadratic in a product form and finding solutions that make each factor zero.

Understanding how to find roots is critical because it aids in analyzing and predicting the behavior of functions. For example, the roots of polynomials signify key turning points and intercepts relevant to graphing functions accurately.