Problem 112 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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-2: gof is twice differentiable at \(x=0\). (a) Statement- 1 is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement-1. (b) Statement- 1 is true, Statement- 2 is false. (c) Statement- 1 is false, Statement- 2 is true. (d) Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1.

Step-by-Step Solution

Answer

(b) Statement- 1 is true, Statement- 2 is false.

Step 1: Understand the Functions

The function \( f(x) = x|x| \) is equal to \( x^2 \) for \( x \geq 0 \) and \( -x^2 \) for \( x < 0 \). The function \( g(x) = \sin x \) is a well-known trigonometric function. We need to check the differentiability of the composition \( gof \), meaning \( g(f(x)) = \sin(x |x|) \), specifically at \( x=0 \).

Step 2: Evaluate f at x=0

For \( x = 0 \), \( f(x) = x|x| = 0 \). Therefore, \( gof(x) = g(f(x)) = g(0) = \sin(0) \). Thus, \( gof(0) = 0 \).

Step 3: Check Differentiability at x=0

To check if \( gof \) is differentiable at \( x = 0 \), compute \( gof'(x) \) as the derivative exists if the limit \( \lim_{x \to 0} \frac{g(f(x))-g(f(0))}{x-0} \) exists. Consider left and right limits. Since \( g(x) = \sin x \) is smooth and differentiable everywhere, only the potential issues with \( f(x) = x|x| \) around \( x = 0 \) need checking. \( f(x) = x|x| \) is continuous and differentiable at \( x=0 \), hence \( g(f(x)) \) is also differentiable at 0.

Step 4: Compute gof'(x) at x=0

Calculate \( gof'(x) = \cos(x |x|) \cdot \frac{d}{dx}(x |x|) \). Since \( f(x) \) is differentiable everywhere, especially at 0, \( gof'(0) = \cos(0) \cdot 0 = 0 \). Thus, \( gof \) is differentiable at 0 and \( gof'(0) = 0 \), which is continuous.

Step 5: Assess Twice Differentiability

For twice differentiability, \( gof \) must have a continuous second derivative at \( x=0 \). Compute \( gof''(x) \) by differentiating \( gof'(x) = \cos(x|x|)\cdot f'(x) \) further. Given changes around \( x=0 \) due to \( f(x) \), the second derivative may not exist continuously. Hence, \( gof \) is not twice differentiable at 0.

Key Concepts

Composite Functions

Composite functions occur when one function is applied inside another. In our exercise, we have the functions \(f(x) = x|x|\) and \(g(x) = \sin x\), with their composition given by \(g(f(x)) = \sin(x|x|)\). Understanding how composite functions work is essential for analyzing their behavior, like differentiability.
To form a composite function, follow these steps:

• Apply the inner function: Compute \(f(x)\).
• Substitute the result into the outer function: Evaluate \(g(f(x))\).

In this case, the key point was figuring out the expression \(g(f(x))\) and considering how \(f(x)\) behaves at \(x = 0\). Composite functions can introduce complexities in properties like differentiability, as they combine different behaviors from each function.

Continuity

Continuity is a fundamental attribute of functions where they do not have abrupt changes or gaps. It means the value of the function approaches the same number from both sides as it gets close to a point.
The function \(f(x)=x|x|\) is given by \(x^2\) for \(x \geq 0\) and \(-x^2\) for \(x < 0\). At \(x = 0\), \(f(x) = 0\) from both sides, demonstrating continuity.
Additionally, the composition \(g(f(x))=\sin(x|x|)\) is continuous at \(x=0\) because \(\sin(x)\) is smooth for all \(x\). Continuity of \(g\) at \(f(x)=0\) supports the overall smoothness of the composite function near \(x=0\).

• Continuity is pivotal for differentiability. If a function isn't continuous at a point, it can't be differentiable there.
• Smooth, continuous trigonometric functions like \(\sin(x)\) often enhance overall continuity when part of a composition.

Twice Differentiability

Twice differentiability means a function has not only a first derivative but also a second derivative that is continuous at a specific point. It's a step further from simple differentiability, giving insight into the function's curvature.
In our problem, \(g(f(x)) = \sin(x|x|)\) is differentiable at \(x = 0\). This means the first derivative \(gof'(x)\) exists and is continuous there, but to be twice differentiable, we need the second derivative \(gof''(x)\) to also exist and be continuous.
However, due to the nature of \(f(x)=x|x|\), especially near 0, the second derivative isn't continuous. The changes in direction in \(f(x)\) around 0 hinder the continuity of \(gof''(x)\). Thus, \(g(f(x))\) isn't twice differentiable at \(x = 0\).

• First differentiability focuses on the tangent's slope.
• Second differentiability examines how this slope changes, requiring more consistency.
• For twice differentiability to be considered, all function components must smoothly transition through all required orders.