Problem 242 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find \(a \& b\) such that the function \(f(x)=a x^{2}-b, \quad|x|<1\) \(=-\frac{1}{|x|}, \quad|x| \geq 1\) is continuous and differentiable function.

Step-by-Step Solution

Answer

The function \(f(x)\) is neither continuous nor differentiable at \(x = 1\) and \(x = -1\). Thus, there are no values of 'a' and 'b' such that the given function will be both continuous and differentiable.

Step 1: Check for Continuity

The function must be equal at \(x = 1\) and \(x = -1\) for it to be continuous. So, firstly, set \(x\) to 1 for both sides of the function. This gives us two equations: \[a(1)^2 -b = - \frac{1}{1}\] and \[a(-1)^2 -b = - \frac{1}{-1}\]. After simplifying these two equations, we get: \(a - b = -1\), and \(a - b = 1\). It can be observed that the two solutions are contradictory, therefore there is no 'a' and 'b' that would satisfy the conditions for continuity.

Step 2: Determining Differentiability

Since our function, \(f(x)\), is not continuous at \(x = 1\) and \(x = -1\), it will automatically fail to be differentiable at these points because a prerequisite of a function's differentiability at a point is that it must be continuous at that point. Hence, it's not necessary to compute the derivative and then check for differentiability.

Key Concepts

Differentiability

Differentiability is a key concept in calculus that tells us whether a function has a derivative at a particular point. Put simply, a function is differentiable at a point if it is smooth and has no sharp corners or discontinuities at that point. For a function to be differentiable, it must satisfy the following conditions:

• The function must be continuous at the point.
• The function must not have any sharp points or cusps at that point.

To determine differentiability mathematically, one often looks at the derivative of the function. If a derivative exists at that point, the function is said to be differentiable there. However, if any point of the function shows a jump or break, it leads to failure in differentiability at that point.

In our case, since continuity at the points where \(x = 1\) and \(x = -1\) was not found, the differentiability of the piecewise function cannot be established either. Hence, it underlines the importance of checking continuity before differentiability when solving such problems.

Piecewise Functions

Piecewise functions are fascinating because they allow different rules for different parts of their domain. They are often used to model real-world scenarios where a uniform behavior doesn’t apply. The general form is written in segments based on specific conditions for the variable.

• A single function can behave distinctly within certain ranges.
• Boundary points are crucial because they determine the transitions between different parts.
• Continuity at boundary points is vital to ensure the function doesn’t 'jump'.

For a piecewise function like the given example, analyzing its continuity and differentiability is essential. Checking continuity ensures that the function remains unbroken through its defined range. In the solution, continuity failed at the borders of \(x = 1\) and \(x = -1\), making it evident that no \(a\) and \(b\) make the function continuous, thus affecting its differentiability.

Quadratic Functions

Quadratic functions are a type of polynomial that are represented in the form \(f(x) = ax^2 + bx + c\). With their characteristic parabolic graph, these functions are widely used in a variety of applications, from physics to finance. Important properties include:

• The vertex, which gives the maximum or minimum point of the function.
• The axis of symmetry, which is a vertical line that passes through the vertex.
• The roots or x-intercepts, which are where the function crosses the x-axis.

In the original exercise, part of the function was quadratic, defined specifically as \(f(x) = ax^2 - b\). When dealing with quadratic functions within a larger piecewise context, it is crucial to ensure that any transitions between function pieces still satisfy key properties like continuity and differentiability. Unfortunately, for this exercise, the values of \(a\) and \(b\) could not be determined to make the piecewise function both continuous and differentiable.