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

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 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

(B) Both statements are true; Statement-2 is not a correct explanation for Statement-1.

Step 1: Understand the Functions

The function \( f(x) = x|x| \) is defined as \( f(x) = x^2 \) when \( x \geq 0 \) and \( f(x) = -x^2 \) when \( x < 0 \). The function \( g(x) = \sin x \) is a trigonometric sine function.

Step 2: Find gof(x)

We need to find \( (g \circ f)(x) = g(f(x)) \). Thus, \((g \circ f)(x) = \sin(x|x|) = \sin(x^2) \) when \( x \geq 0 \) and \((g \circ f)(x) = \sin(-x^2) = \sin(x^2)\) when \( x < 0 \). Therefore, \((g \circ f)(x) = \sin(x^2)\) for all \(x\).

Step 3: Differentiability of gof at x=0

To check differentiability at \( x=0 \), find the derivative \( (g \circ f)'(x) \). The derivative is \( (g \circ f)'(x) = 2x\cos(x^2) \) using the chain rule. Evaluate this at \( x=0 \), giving \( 0\cos(0) = 0 \). Therefore, \( g \circ f \) is differentiable at \( x=0 \).

Step 4: Continuity of the Derivative at x=0

The derivative \( 2x\cos(x^2) \) is continuous around \( x=0 \) since it involves continuous functions. Thus, the derivative of \( g \circ f \) is continuous at \( x=0 \). Statement-1 is true.

Step 5: Twice Differentiability of gof at x=0

Find the second derivative: differentiate \( (g \circ f)'(x) = 2x\cos(x^2) \) again. Using product and chain rules, we get \( (g \circ f)''(x) = 2\cos(x^2) - 4x^2\sin(x^2) \). Evaluate at \( x=0 \), which results in \( 2 \cdot 1 - 0 = 2 \). Thus, \( g \circ f \) is twice differentiable at \( x=0 \).

Step 6: Analyze Statement-2

Statement-2 is true as \( g \circ f \) being twice differentiable implies it is differentiable, fulfilling the condition of Statement-1. However, it does not provide additional explanation for Statement-1's continuity reasons.

Key Concepts

Chain Rule

The chain rule is a crucial technique in calculus for finding the derivative of a composite function. When you have two functions, say \'f\' and \'g\', and you need to differentiate the composition of these functions, denoted as \( g \, \circ \, f \), you apply the chain rule. It states that the derivative of \( g(f(x)) \) is the derivative of \( g \) evaluated at \( f(x) \) multiplied by the derivative of \( f(x) \).

In our original exercise, we used the chain rule to differentiate \( (g \, \circ \, f)(x) = \sin(x^2) \). The derivative \( (g \, \circ \, f)'(x) \) was computed as \( 2x\cos(x^2) \). Here, \( f(x) = x^2 \) and its derivative is \( 2x \), while for \( g(x) = \sin x \), the derivative is \( \cos x \). Applying the chain rule seamlessly provided us the differentiability at \( x=0 \).

The chain rule highlights the interconnectedness of functions and how their rates of change relate when composed. This rule is indispensable for dealing with complex combinations like \( \sin(x^2) \) and ensures a deeper understanding of functional behavior through derivatives.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, play a foundational role in mathematics, especially within calculus. These functions define relationships between the angles and sides of right-angled triangles, but they extend beyond geometry into modeling periodic phenomena in physics, engineering, and more.

In our exercise, the function \( g(x) = \sin x \) is a trigonometric function that acts upon \( f(x) = x|x| \). When composed, it creates \( g(f(x)) = \sin(x^2) \). This highlights the ability of trigonometric functions to wrap or 'oscillate' other functions within their cyclical patterns.

Trigonometric functions are continuous everywhere and have derivatives that are well-known: \( \sin x \) differentiates to \( \cos x \), while \( \cos x \) differentiates to \(-\sin x \). This continuity and smoothness make them incredibly useful in solving differentiation problems like the one encountered with \( (g \, \circ \, f)(x) \) in the exercise.

Continuity

Continuity is a property of a function that ensures there are no 'jumps' or 'gaps' in its graph. A function \( g(x) \) is continuous at a point \( x = a \) if \( \lim_{x \to a} g(x) = g(a) \). Essentially, you can draw the function's graph without lifting your pencil at that point.

In the context of differentiability, continuity of a derivative is often considered. For a function to be differentiable at a point, it must be continuous there, but a derivative being continuous can offer insights into the behavior of derivative functions, as was evaluated at \( x = 0 \) in our original exercise.

When we determined the continuity of the derivative \( 2x\cos(x^2) \) at \( x = 0 \), we saw no disruption, confirming smooth and consistent behavior through and around that point. This bolsters the argument behind statement-1 in the exercise, ensuring the derivative's reliability and functionality at \( x = 0 \).

Twice Differentiability

A function is twice differentiable if it has a second derivative everywhere within its domain. This property is essential in calculus because it gives insight into the function's concavity and the rate of change of its slope (i.e., how the function bends).

In our exercise, checking twice differentiability involves calculating the second derivative of \( (g \, \circ \, f)(x) = \sin(x^2) \). We found this to be \( (g \, \circ \, f)''(x) = 2\cos(x^2) - 4x^2\sin(x^2) \), which showed that at \( x = 0 \), the value was \( 2 \), indicating the second derivative exists and is well-defined there.
The concept not only confirms statement-2 but expands upon statement-1's truthfulness by ensuring there's not just continuity and differentiability, but also smoothness of change in the derivative itself, strengthening the argument for a comprehensive understanding of the function's behavior at \( x = 0 \).