Problem 39 - A Complete Resource Book in Mathematics for JEE Main

Question

The values of constants \(a\) and \(b\) so as to make the function \(f(x)=\left\\{\begin{array}{l}\frac{1}{|x|},|x| \geq 1 \\ a x^{2}+b,|x|<1\end{array}\right.\) continuous as well as differentiable for all \(x\), are (A) \(a=\frac{-1}{2}, b=\frac{3}{2}\) (B) \(a=\frac{1}{2}, b=\frac{3}{2}\) (C) \(a=\frac{-1}{2}, b=\frac{-3}{2}\) (D) None of these

Step-by-Step Solution

Answer

(A) \(a = -\frac{1}{2}, b = \frac{3}{2}\).

Step 1: Identify Points of Interest

The function has different expressions based on the value of \(|x|\). We need to particularly check the points where the expression changes, which are at \(x = 1\) and \(x = -1\).

Step 2: Check Continuity at x = 1

For the function to be continuous at \(x = 1\), the limit as \(x\) approaches 1 from both sides and the value of the function at \(x = 1\) should be equal. Calculate \(\lim_{x \to 1^-}(ax^2 + b)\). Since \(|x| < 1\) applies, this gives \(a(1)^2 + b = a + b\). The function at \(x = 1\) is \(\frac{1}{|1|} = 1\). Thus, for continuity, \(a + b = 1\).

Step 3: Check Differentiability at x = 1

The function is differentiable if its derivative from both sides at \(x = 1\) is equal. The derivative of \(f(x)\) when \(|x| \ge 1\) is \(-\frac{1}{x^2}\). At \(|x| < 1\), \(f'(x) = 2ax\). The left-hand derivative at \(x = 1\) is \(2a \cdot 1 = 2a\) and the right-hand derivative is \(-\frac{1}{(1)^2} = -1\). Equating these gives \(2a = -1\), hence \(a = -\frac{1}{2}\).

Step 4: Solve for Constant b

With \(a = -\frac{1}{2}\), substitute in the equation \(a + b = 1\) found in Step 2. \(-\frac{1}{2} + b = 1\) gives \(b = \frac{3}{2}\).

Step 5: Verify at x = -1

Repeat the check for continuity and differentiability at \(x = -1\). Similar reasoning applies due to symmetry: \(a + b = 1\) from continuity and \(2a = -1\) from differentiability already satisfy these at \(x = -1\). Thus, no further changes are needed.

Key Concepts

Piecewise Functions

Piecewise functions are a type of function that have different expressions for different intervals of the input variable, often denoted as \(x\). These are particularly useful when the behavior of a function changes based on the value of \(x\). For the given function \(f(x)\), it is defined piecewise as:- \(f(x) = \frac{1}{|x|}\) when \(|x| \geq 1\)- and \(f(x) = ax^2 + b\) when \(|x| < 1\).This function has a "piece" of expressions for certain domains of \(x\), creating points of interest at \(x = 1\) and \(x = -1\), where the expression formulas switch. Understanding piecewise functions is crucial because each "piece" can behave very differently, meaning calculations for continuity and differentiability must be handled separately for each interval before unifying the overall behavior.

Limits

Limits help us understand the behavior of functions as they approach a particular point. For a piecewise function to be continuous, the limits from both sides of the point where the function changes its expression must be the same. For this problem, consider checking continuity at the crucial point \(x = 1\):- Calculate \( \lim_{x \to 1^-} (ax^2 + b) \), which simplifies to \(a + b\), since \(|x| < 1\) applies as \(x\) approaches from the left.- For continuity at \(x = 1\), set \(a + b = f(1) = 1\), where \(f(1) = \frac{1}{|1|} = 1\).This equality ensures there's no "jump" between the two pieces of our piecewise function. Similarly, a limit check at \(x = -1\) would confirm that the other critical point behaves seamlessly.

Derivatives

Derivatives provide information about the rate of change of a function, and are necessary to evaluate differentiability, particularly for piecewise functions. These are important at points where the pieces of the function join.In this exercise, for the function to be differentiable at \(x = 1\), the derivative must be the same from both sides of the point.- For \(|x| < 1\), the derivative is \(f'(x) = 2ax\).- For \(|x| \geq 1\), the derivative is \(f'(x) = -\frac{1}{x^2}\).Equating these at \(x = 1\):- The left-hand derivative is \(2a\),- while the right-hand derivative, from \(|x| \geq 1\), is \(-1\).Equating gives: \(2a = -1\), solving for \(a\) results in \(a = -\frac{1}{2}\). This ensures the slopes match where the pieces meet, maintaining smoothness at the transition.

Symmetry in Functions

Symmetry in a function can simplify the calculation process, as it often implies that the function behaves the same under certain transformations. In this problem, symmetry around the y-axis (even symmetry) holds since the function's expression for \(|x|\) mirrors between its positive and negative counterparts.For the function \(f(x)\), evaluate around symmetry:- Points of interest \(x = 1\) and \(x = -1\) require the same treatment with continuity and differentiability.By testing these points symmetrically:- The same conditions \(a + b = 1\) and \(2a = -1\) apply.Thus, knowing a function is even or symmetrical can help confirm solutions more quickly across specified limits, offering a consistent approach without redundant calculations. This understanding was encapsulated in the provided solution to ensure functionality across the full domain of the piecewise function.