Problem 48 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=x|x|\) and \(g(x)=\sin x\). Statement-1: gof is differentiable at \(x=0\) and its derivative is continuous at that point. Statement- \(\mathbf{2}\) : gof is twice differentiable at \(x=0\). (A) Statement-1 is true, Statement- 2 is true; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is true, Statement-2 is true; Statement- 2 is not a correct explanation for Statement- 1 (C) Statement- 1 is true, Statement- 2 is false (D) Statement- 1 is false, Statement- 2 is true

Step-by-Step Solution

Answer

C: Statement-1 is true, Statement-2 is false.

Step 1: Understanding the composition

We have two functions, \( f(x) = x|x| \) and \( g(x) = \sin x \). The function \( f(x) \) can be rewritten as: \( f(x) = x^2 \text{ for } x \geq 0 \) and \( f(x) = -x^2 \text{ for } x < 0 \). This means \( f(x) \) is continuous but not differentiable at \( x = 0 \) since it has a cusp there. The composition \( g\circ f = g(f(x)) = \sin(f(x)) \).

Step 2: Checking differentiability at x=0

Since \( f(x) \) is not differentiable at \( x = 0 \), we need to be cautious. Calculate \( g(f(x)) = \sin(f(x)) \) around \( x = 0 \). Since \( f(x) = x|x| \approx 0 \) near \( x = 0 \), \( g(f(x)) = \sin(0) = 0 \). Hence, \( g(f(x)) = 0 \) for very small \( x \), meaning it behaves well near \( x = 0 \). Therefore, \( g\circ f \) is differentiable at \( x = 0 \) and its derivative \( (g\circ f)'(0) \) is also 0 since \( \left. \frac{d}{dx} \sin(0) \right|_{x=0} = 0 \). Thus, Statement-1 is true.

Step 3: Checking the continuity of the derivative

For \( (g\circ f)'(x) \) to be continuous at \( x = 0 \), \( \lim_{x \to 0} (g\circ f)'(x) \) must exist and equal \( (g\circ f)'(0) = 0 \). Since \( g(f(x)) = 0 \) continuously for \( x \) near zero, its derivative doesn't change, confirming this condition. Therefore, the derivative is continuous at \( x = 0 \).

Step 4: Checking twice differentiability at x=0

To determine if \( g\circ f \) is twice differentiable at \( x = 0 \), we need \( (g\circ f)''(0) \) to exist. Since \( f(x) = x|x| \) is not differentiable at \( x = 0 \), \( g\circ f \) cannot be twice differentiable there. Thus, Statement-2 is false.

Key Concepts

Differentiability

Differentiability is a foundational concept in calculus that deals with whether a function has a derivative at each point in its domain. A function is said to be differentiable at a point if it has a well-defined derivative there. This means that the function can be approximated by a linear function at that point, offering a strong sense of smoothness. For example, in the function composition described in the problem, the function \( f(x) = x|x| \) is differentiable everywhere except at \( x = 0 \) because it has a cusp at that point.

When examining composite functions, like \( g(f(x)) = \sin(f(x)) \), differentiability at a specific point like \( x = 0 \) becomes critical to understand. Despite \( f(x) \) not being differentiable at \( x = 0 \), the surrounding behavior of \( g(f(x)) \) near this point allows \( g \circ f \) to be handled smoothly. Since \( \sin(0) = 0 \), the composition \( g(f(x)) \) approaches zero smoothly as \( x \) approaches zero, thus allowing \( g \circ f \) to be differentiable at \( x = 0 \).

This result illustrates an interesting scenario where the differentiability of a composition can differ from its individual components. Keep in mind that a function's differentiability often requires checking both the original functions and their compositions, especially at crucial points like cusps or discontinuities.

Continuity of Derivative

The continuity of a derivative is an extension of the basic concept of continuity. A derivative is said to be continuous at a point if it does not have any abrupt changes around that point, meaning the limit of the derivative as you approach the point is equal to the derivative at that point. In essence, it's about the smoothness of the rate of change itself.

For the composite function \( g \circ f \), where \( g(x) = \sin x \) and \( f(x) = x|x| \), ensuring continuity of the derivative at \( x = 0 \) involves checking whether \( \, \lim_{x \to 0} \, (g\circ f)'(x) \) matches \( (g \circ f)'(0) \). This provides us confidence that there isn't a sudden jump or spike in the behavior of the derivative around \( x = 0 \).

In the solution, since \( g(f(x)) = \sin(f(x)) \) remains zero continuously near zero, the derivative \( (g \circ f)'(x) \) doesn't exhibit any discontinuity. Thus the function's derivative behaves well around the point, and continuity is assured at \( x = 0 \). This helps confirm that Statement 1 is true as well, allowing smoothness not just in the function itself, but its rate of change as well.

Twice Differentiable

A function is twice differentiable if it has a second derivative at every point of interest. This means the function is smooth enough to have a derivative and that this derivative also has a derivative at those points.

However, in the provided problem, the twice differentiability of \( g \circ f \) at \( x = 0 \) doesn't hold. The function \( f(x) = x|x| \) lacks differentiability at \( x = 0 \), meaning that \( (g \circ f)'(x) \), the first derivative, might introduce an abrupt or undefined behavior when attempting to compute \( (g \circ f)''(x) \).

Such irregularities at a point like \( x = 0 \) suggest that there’s not enough smoothness to ensure the existence of a second derivative. Therefore, in this case, \( (g \circ f)(x) \) is not twice differentiable at \( x = 0 \), making Statement 2 false. This conclusion highlights how twice differentiability imposes stricter conditions than simple differentiability, often requiring both the function and its first derivative to be smooth and continuous.