Problem 6 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f(x)=5-|x-2|\) and \(g(x)=|x+1|, x \in \mathrm{R}\). If \(f(x)\) attains maximum value at \(\alpha\) and \(g(x)\) attains minimum value at \(\beta\), then \(\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^{2}-5 x+6\right)}{x^{2}-6 x+8}\) is equal to : [April 12, 2019 (II)] (a) \(1 / 2\) (b) \(-3 / 2\) (c) \(-1 / 2\) (d) \(3 / 2\)

Step-by-Step Solution

Answer

The limit evaluates to \(\frac{1}{2}\), option (a).

Step 1: Determine the maximum value of f(x)

The function is given by \(f(x) = 5 - |x - 2|\). The absolute value function, \(|x - 2|\), has a minimum value of \(0\) when \(x = 2\). Thus, \(f(x)\) attains its maximum value of 5 at \(\alpha = 2\).

Step 2: Determine the minimum value of g(x)

The function is given by \(g(x) = |x + 1|\). The absolute value function, \(|x + 1|\), has a minimum value of \(0\) when \(x = -1\). Thus, \(g(x)\) attains its minimum value at \(\beta = -1\).

Step 3: Calculate the expression for the limit given \(\alpha\) and \(\beta\)

Given that \(\alpha = 2\) and \(\beta = -1\), we have \(-\alpha\beta = -2 * (-1) = 2\). We need to evaluate the limit: \[\lim_{x\rightarrow 2} \frac{(x-1)(x^2-5x+6)}{x^2-6x+8}\] The expression simplifies as below.

Step 4: Factorize and simplify expressions

Factor the quadratic expressions: - \(x^2-5x+6 = (x-2)(x-3)\) - \(x^2-6x+8 = (x-2)(x-4)\) This allows us to rewrite the limit as: \[\lim_{x\rightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}\]Cancel \((x-2)\) from numerator and denominator:

Step 5: Simplify the cancellables and find the limit

Now, simplifying and using \(x \to 2\) gives:\[\lim_{x\rightarrow 2} \frac{(x-1)(x-3)}{(x-4)} = \frac{(2-1)(2-3)}{2-4} = \frac{1 \times (-1)}{-2} = \frac{1}{2}\]

Step 6: Conclusion from the simplification

The resulting value from the simplification is \(\frac{1}{2}\). Therefore, the answer to the limit is \(\frac{1}{2}\).

Key Concepts

Piecewise Functions

A piecewise function is a single function defined by multiple sub-functions, each applied to a certain interval of the function's domain. This makes them especially useful to represent equations that have different outputs depending on the input value, such as those defined by conditions or thresholds. In the context of our problem, both functions, \(f(x) = 5 - |x - 2|\) and \(g(x) = |x + 1|\), can be viewed as piecewise functions due to the absolute value. Remember that when dealing with piecewise functions, it’s essential to consider the behavior of each segment separately.

Limits in Calculus

The concept of limits is integral in calculus and is fundamental in defining derivatives and integrals. Limits help us understand the behavior of functions as they approach a certain point. In this exercise, the limit \(\lim_{x \rightarrow -\alpha\beta} \frac{(x-1)(x^2-5x+6)}{x^2-6x+8}\) practically evaluates the behavior of the function output as \(x\) approaches \(2\), after finding \(-\alpha\beta = 2\). Understanding limits is key to assessing the continuity and differentiability of functions.

Problem-solving Strategies

Navigating math problems like this one requires a strategic approach. Here are some effective strategies:

• Clearly identify what needs to be solved, such as determining \(\alpha\) and \(\beta\) by analyzing the maximum and minimum values of given functions.
• Simplify complex looking expressions by factoring whenever possible to manage calculations easily.
• Apply limits only after canceling indeterminacies to avoid incorrect answers and unnecessary complications.

These problem-solving strategies can simplify the seemingly complicated steps into manageable chunks.

Factorization

Factorization is the process of breaking down expressions into simpler elements called factors. This is useful in simplifying equations and finding roots of polynomial expressions. In our solution, both quadratic expressions \(x^2-5x+6\) and \(x^2-6x+8\) were factorized into products of linear factors:

• \(x^2-5x+6 = (x-2)(x-3)\)
• \(x^2-6x+8 = (x-2)(x-4)\)

This factorization was critical to simplify the original limit expression and eliminate common terms. By canceling these out, we can focus directly on evaluating the limit.

Absolute Value Functions

Absolute value functions, denoted by \(|x|\), give the distance of a number from zero on a number line, disregarding direction. Hence, \(|x - 2|\) in \(f(x)\) and \(|x + 1|\) in \(g(x)\) measure the deviations from 2 and -1 respectively. The minimum value for any absolute function is zero, occurring at the point where the expression inside the absolute value equals to zero. Therefore, knowing this property helps find the extreme values of the functions: \(f(x)\)'s maximum and \(g(x)\)'s minimum. This understanding is critical in determining key values like \(\alpha\) and \(\beta\) in our problem.