Problem 13 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{a}, \mathrm{b} \in \mathrm{R},(\mathrm{a} \neq 0)\). if the function \(\mathrm{f}\) defined as \(f(x)= \begin{cases}\frac{2 x^{2}}{a} & , 0 \leq x<1 \\ a & , 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{3}} & , & \sqrt{2} \leq x<\infty\end{cases}\) is continuous in the interval \([0, \infty)\), then an ordered pair \((\mathrm{a}, \mathrm{b})\) is: (a) \((-\sqrt{2}, 1-\sqrt{3})\) (b) \((\sqrt{2},-1+\sqrt{3})\) (c) \((\sqrt{2}, 1-\sqrt{3})\) (d) \((-\sqrt{2}, 1+\sqrt{3})\)

Step-by-Step Solution

Answer

The ordered pair is \((\sqrt{2}, 1-\sqrt{3})\).

Step 1: Continuity Condition Check at x=1

For the function to be continuous at \(x = 1\), the left-hand and right-hand limits and the value at \(x = 1\) must all be equal. Evaluate the left limit as \(x\) approaches 1:\[\lim_{{x \to 1^-}} f(x) = \frac{2 \cdot 1^2}{a} = \frac{2}{a}.\]The value at \(x = 1\):\[f(1) = a.\]Set them equal for continuity:\[\frac{2}{a} = a.\]Solve this equation: \(a^2 = 2 \rightarrow a = \pm \sqrt{2}.\)

Step 2: Continuity Condition Check at x=√2

Check continuity at \(x = \sqrt{2}\). First, compute the left-hand limit as \(x\) approaches \(\sqrt{2}\) from the left:\[\lim_{{x \to \sqrt{2}^-}} f(x) = a.\]The function value from the given function definition is also \(a\). Now find the right-hand limit:\[\lim_{{x \to \sqrt{2}^+}} f(x) = \frac{2b^2 - 4b}{(\sqrt{2})^3} = \frac{2b^2 - 4b}{2 \sqrt{2}} = \frac{b^2 - 2b}{\sqrt{2}}.\]For continuity, equate this to \(a\):\[\frac{b^2 - 2b}{\sqrt{2}} = a.\]Substitute \(a = \pm \sqrt{2}\):\[b^2 - 2b = 2 \, or \, -2.\]Solve each case for \(b\).

Step 3: Solve b when a=√2

Substitute \(a = \sqrt{2}\) into the equation:\[b^2 - 2b = 2.\]Rearrange and solve the quadratic equation \(b^2 - 2b - 2 = 0\).Use the quadratic formula:\[b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}.\]The values for \(b\) are \(1 + \sqrt{3}\) or \(1 - \sqrt{3}\). Thus, the valid ordered pair is \((\sqrt{2}, 1-\sqrt{3})\).

Step 4: Verify other options

If \(a = -\sqrt{2}\), substitute into the continuity equation at \(x = \sqrt{2}\):\[b^2 - 2b = -2,\]which simplifies to \(b^2 - 2b + 2 = 0\). Solve this using the quadratic formula:\[b = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2} = \frac{2 \pm \sqrt{-4}}{2},\]which results in no real solution. Therefore, \(a = -\sqrt{2}\) does not give a valid real number \(b\) for continuity.

Key Concepts

Piecewise Functions

Piecewise functions are a type of function defined by different expressions depending on the interval of the input variable. These functions are essential in mathematical modeling where different rules apply under varying conditions.
Understanding how these functions work involves analyzing each part separately:

• Firstly, recognize which piece of the function applies to your input. In our given exercise, depending on the value of \(x\), whether it's between \([0, 1)\), \([1, \sqrt{2})\), or \([\sqrt{2}, \infty)\), we use different expressions.
• Evaluate these expressions by taking the constraints in the input into account. You might be using a simple format in some intervals and a more complex, like a fraction, in others.
• Check continuity, which means ensuring that not only do the separate equations match at their boundaries, but also that they connect smoothly without any jumps or gaps.

Piecewise functions give flexibility in mathematical descriptions, especially useful in physics, economics, and computer science where conditions change outcomes.

Limits in Calculus

The concept of limits forms the backbone of calculus. It's about approaching a value as close as needed. In the exercise, limits examine the behavior of a piecewise function at certain points to ensure continuity.
Understanding limits involves:

• Left-hand and right-hand limits: At any point, consider approaching it from either side - left and right. For continuity, these must be equal.
• Evaluating a function as it approaches a certain point: Compute the values from different pieces of the function as the input nears a boundary.

In calculus, limits help us transition smoothly between values and ensure that functions make logical sense in every section. In practical applications, they explain behavior that's complex to compute directly, like changing speeds or economic models in varying conditions.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically in the form \(ax^2 + bx + c = 0\). They are crucial in various mathematical and real-world scenarios.
In this context, to solve for continuity, we dive into solving quadratic equations. Here's how it works:

• Identifying coefficients: From our step-by-step solution, coefficients might change based on conditions applied by piecewise functions.
• Quadratic formula: We use \(b = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find possible values of \(b\) in our exercise.

The quadratic equations unravel potential solutions for variables in our piecewise function, serving practical purposes in finding dimensions, cost minimization, or trajectory calculations.