Problem 81 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=\left\\{\begin{array}{cl}\frac{1}{|x|} & |x| \geq 1 \\ a x^{2}+b & |x|<1\end{array}\right.\) If \(f\) is continuous and differentiable at every point, then (A) \(a=\frac{1}{2}\) (B) \(a=-\frac{1}{2}\) (C) \(b=\frac{3}{2}\) (D) \(b=\frac{-3}{2}\)

Step-by-Step Solution

Answer

(B) \(a = -\frac{1}{2}\) and (C) \(b = \frac{3}{2}\) are correct.

Step 1: Understand the Structure of the Function

The function \( f(x) \) is defined piecewise; it changes its formula depending on whether \(|x| \geq 1\) or \(|x| < 1\). For \(|x| \geq 1\), \(f(x) = \frac{1}{|x|}\). For \(|x| < 1\), \(f(x) = ax^2 + b\). We need to ensure both continuity and differentiability at the point where these pieces meet, specifically at \(x = 1\) and \(x = -1\).

Step 2: Check Continuity at x = 1

For the function to be continuous at \(x = 1\), the limits from the left and right must equal the function's value at \(x = 1\). \(f(1) = \frac{1}{1} = 1\). The left-hand limit \(\lim_{x \to 1^-} (ax^2 + b) = a \cdot 1^2 + b = a + b\). Set this equal to \(f(1)\): \[ a + b = 1. \]

Step 3: Check Continuity at x = -1

Similarly, for the function to be continuous at \(x = -1\), the limits must match the function value. \(f(-1) = \frac{1}{1} = 1\). The left-hand limit \(\lim_{x \to -1^+} (ax^2 + b) = a \cdot (-1)^2 + b = a + b\). Set this equal to \(f(-1)\): \[ a + b = 1. \]

Step 4: Differentiate and Assess Differentiability at x = 1 and x = -1

For differentiability, ensure the derivatives from both sides match. The derivative of \(\frac{1}{|x|}\) for \(x > 0\) is \(-\frac{1}{x^2}\). For \(ax^2 + b\), the derivative is \(2ax\). At \(x = 1\), \(\lim_{x \to 1^-} (2ax) = 2a\). Equate this to the derivative from the other side: \[-\frac{1}{1^2} = 2a\] \[a = -\frac{1}{2}.\]

Step 5: Find b Using the Value of a

Substitute \(a = -\frac{1}{2}\) into the equation from continuity at \(x = 1\) and \(x = -1\): \[-\frac{1}{2} + b = 1.\] Solving for \(b\), \[b = \frac{3}{2}.\]

Step 6: Conclusion

The function is both continuous and differentiable at every point only if \(a = -\frac{1}{2}\) and \(b = \frac{3}{2}\). Thus, statements (B) and (C) are true.

Key Concepts

Understanding Continuity in Piecewise Functions

Continuity in mathematics ensures that a function behaves predictably; there are no sudden jumps or gaps. For a piecewise function like our given one, continuity must hold at the points where the function switches its definitions.

For the function \( f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \ ax^2 + b, & |x| < 1 \end{cases} \), we must check continuity at the boundaries \( x = 1 \) and \( x = -1 \). These are the points where the function's rule changes.

To be continuous at these points, the limits from both the left and right of the boundary positions must match the actual value of the function at these points. Here are the steps to ensure continuity:

• Calculate the value of the function at the boundary points.
• Determine the left-hand limit (from the \( |x| < 1 \) side) approaching each boundary.
• Match the left-hand limit to the function's value at the boundary.

For \( x = 1 \), the function yields \( f(1) = 1 \), with the left-hand limit being \( a + b \). Therefore, \( a + b = 1 \) assures continuity at \( x = 1 \). Similarly, do the procedure for \( x = -1 \). It's crucial the limits equal \( f(x) \) at the points of potential discontinuity.

Differentiability of Piecewise Functions

Differentiability is a concept that elevates continuity by ensuring not only does a function make smooth transitions but also does so with a constant rate of change, devoid of any sharp corners or cusps.For differentiability at a point, the left-hand derivative must equal the right-hand derivative at that point. In our piecewise function, differentiability must be checked at \( x = 1 \) and \( x = -1 \) where the definition of the function changes.
Here's what we look at:

• The derivative of \( \frac{1}{|x|} \), when moving from the positive to negative x and vice versa.
• The derivative of \( ax^2 + b \), specifically as it approaches these boundary points.
• Equating these derivatives from both sides of the discontinuous potential, to find constants \(a\) and \(b\).

At \( x = 1 \):- The derivative \( \frac{d}{dx} \left( \frac{1}{x} \right) \) when \( x > 0 \) is \( -\frac{1}{x^2} \).- The derivative \( \frac{d}{dx} (ax^2 + b) \) is \( 2ax \).For differentiability, \( 2a \cdot 1 = -\frac{1}{1^2} \), simplifying to find \( a = -\frac{1}{2} \).By analyzing the rates of change at these critical points, we've determined the constraints on the function constants \(a\) and \(b\) to ensure smoothness throughout the piecewise structure.

The Role of Limits in Piecewise Functions

Limits are foundational in understanding how functions behave as they approach specific points. For piecewise functions, ensuring the limits from different piece definitions meet requirements is key to analyzing continuity and differentiability.In our scenario, we're particularly interested in how:

• The function approaches from different segments as \( x \to 1^- \) and \( x \to 1^+ \).
• It behaves similarly for \( x \to -1^- \) and \( x \to -1^+ \).

For continuity at \( x = 1 \), it's required that:- \( \lim_{x \to 1^-} (ax^2 + b) = 1 \).- \( a + b = 1 \) from this limit allows the function segments to meet smoothly.For differentiability:- The limits of the derivatives must match. Hence, for left and right derivatives at \( x = 1 \), the constraint \( a = -\frac{1}{2} \) was found.These calculations ensure that as you approach specific points, the behavior of both function values and their derivatives transition without abrupt changes. Calculating limits correctly is an integral part of assessing if a piecewise function will be continuous and differentiable across all its segments.