Problem 39 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The values of constants \(a\) and \(b\) so as to make the function \(f(x)=\left\\{\begin{array}{ll}\frac{1}{|x|},|x| \geq 1 \\ a x^{2}+b,|x|<1\end{array}\right.\) 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

The correct values are \(a = -\frac{1}{2}, b = \frac{3}{2}\).

Step 1: Check Continuity at Critial Points

For the function to be continuous at the critical points (which are points where the definition of the function changes, here it's at \(|x| = 1\)), we must have the limits from both sides coinciding and equal to the function value. Check the value at \( x = 1 \):\[ \lim_{x \to 1^-} f(x) =\lim_{x \to 1^-} (ax^2 + b) = a(1)^2 + b = a + b. \] And:\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{|x|} = \frac{1}{1} = 1. \] For continuity at \( x = 1 \), set them equal: \(a + b = 1\). Similar conditions apply at \( x = -1 \).

Step 2: Check Differentiability at Critical Points

For differentiability at \( x = 1 \), ensure the derivatives from both sides match. Find derivatives:\[ \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{a(1 + h)^2 + b - 1}{h} = \lim_{h \to 0^-} \frac{a(1 + 2h + h^2) + b - 1}{h} = 2a. \] And:\[ \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0^+} \frac{-h}{h(1+h)} = -1. \] Equate the left and right derivatives: \(2a = -1\).

Step 3: Solve Equations

From Step 1, we have \(a + b = 1\). From Step 2, we have \(2a = -1\). Solve these equations: - From \(2a = -1\), it follows \(a = -\frac{1}{2}\). - Substitute \(a = -\frac{1}{2}\) into \(a + b = 1\): \(-\frac{1}{2} + b = 1\). Solve for \(b\): - \(b = 1 + \frac{1}{2} = \frac{3}{2}\).

Step 4: Verify the Continuity and Differentiability

With \(a = -\frac{1}{2}\) and \(b = \frac{3}{2}\), check continuity \(a + b = 1\) and differentiability \(2a = -1\). These conditions are satisfied, confirming the function's continuity and differentiability for these values. Therefore, the constants found ensure the function is both continuous and differentiable.

Key Concepts

Continuity

Understanding continuity in functions, especially piecewise functions, is important in calculus. A function is continuous at a point if there is no interruption in its graph at that point. Essentially, this means that if you pick any point along the curve, you can draw the function without lifting your pencil. For a piecewise function like the one in our exercise, continuity must be checked at the boundaries where the function's definition changes, which are called critical points.
To ensure continuity at a critical point, the left-hand limit, right-hand limit, and the function value at the point must all be the same. In our example, continuity is checked at \(|x| = 1\), meaning both limits as approaching from the positive side (\(x \to 1^+\)) and from the negative side (\(x \to 1^-\)) must equal each other and the value of the function at that point.

• Left-hand limit: \(a + b\) at \(|x| < 1\).
• Right-hand limit: \(1\) at \(|x| \geq 1\).
• For continuity, both limits and the function value should be \(1\).

When these conditions are met, the function is continuous at the point.

Differentiability

Differentiability is a measure of how smoothly a function changes at a given point. If a function has a derivative at a point, it means the function is differentiable at that point. For a function to be differentiable, it also has to be continuous there, but the converse isn't always true.
In our exercise, differentiability is checked at the critical point \(x = 1\), where the derivative should be the same from both sides of that point. By calculating the limit of the difference quotient from the left and right sides, we determine whether the derivative exists at the critical point.
For \(x = 1\):

• Left-hand derivative from the segment \(|x| < 1\): \( \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = 2a \).
• Right-hand derivative from the segment \(|x| \geq 1\): \( \lim_{h \to 0^+} \frac{\frac{1}{1+h} - 1}{h} \to -1 \).

For differentiability, the derivatives from both directions must agree. Matching the left and right derivative gives the equation \(2a = -1\), leading to the solution of \(a\).

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each applying to a specific interval in the domain. This type of function can model complex real-world situations that cannot be described by a single formula. A piecewise function's graph can consist of different shapes that join together at certain points called critical points.
In our exercise, the function definition changes at \(|x| = 1\). - For \(|x| \geq 1\), the function is defined as \( \frac{1}{|x|} \), indicating a hyperbolic behavior.- For \(|x| < 1\), the function takes the form of \(a x^2 + b\), a parabolic curve.Critical points of piecewise functions are where the sub-functions connect. At these points, we ensure both the continuity and differentiability to have a smooth overall function. By carefully setting the constants \(a\) and \(b\), we manage the transitions at \(|x| = 1\) smoothly, making the function seamless in behavior across its entire domain.