Problem 45 - A Complete Resource Book in Mathematics for JEE Main

Question

Amongst the following functions, a function that is differentiable at \(x=0\) is (A) \(\cos (|x|)-|x|\) (B) \(\cos (|x|)+|x|\) (C) \(\sin (|x|)+|x|\) (D) \(\sin (x \mid)-|x|\)

Step-by-Step Solution

Answer

The function \(j(x) = \sin (|x|)-|x|\) is differentiable at \(x=0\).

Step 1: Understanding Differentiability

To determine if a function is differentiable at a point, it must be continuous at that point and its derivative must exist at that point. Hence, we first need to ensure each function is continuous at \(x=0\) and then check if the derivative exists.

Step 2: Check Continuity at \(x=0\)

Evaluate each function to ensure it is continuous at \(x=0\). For each function, replace \(x\) with \(0\) and see if the value from both sides of zero approaches the same limit. This confirms continuity at the point.

Step 3: Function (A) Evaluation

Evaluate \(f(x) = \cos (|x|)-|x|\) at \(x=0\). We find that \(f(0) = \cos(0) - 0 = 1\). The function is symmetric around \(x=0\) due to the absolute values, indicating continuity since both side limits approach 1.

Step 4: Function (B) Evaluation

Evaluate \(g(x) = \cos (|x|)+|x|\) at \(x=0\). We find that \(g(0) = \cos(0) + 0 = 1\). As with (A), both sides of zero approach 1, indicating it is continuous at \(x=0\).

Step 5: Function (C) Evaluation

Evaluate \(h(x) = \sin (|x|)+|x|\) at \(x=0\). We find that \(h(0) = \sin(0) + 0 = 0\). Continuity is preserved since both side limits are 0 as well.

Step 6: Function (D) Evaluation

Evaluate \(j(x) = \sin (|x|)-|x|\) at \(x=0\). We find that \(j(0) = \sin(0) - 0 = 0\). This function is continuous at \(x=0\) as well.

Step 7: Check Differentiability for (A)

Differentiate \(f(x) = \cos (|x|)-|x|\): split into \(x>0\) and \(x<0\) to find left and right derivatives. The derivatives do not match at zero, making it not differentiable at \(x=0\).

Step 8: Check Differentiability for (B)

Differentiate \(g(x) = \cos (|x|)+|x|\): similarly split into \(x>0\) and \(x<0\). Left and right derivatives do not match at zero, indicating \(g(x)\) isn't differentiable at this point either.

Step 9: Check Differentiability for (C)

Differentiate \(h(x) = \sin (|x|)+|x|\): The left-hand derivative is \(1\) and the right-hand derivative is \(-1\), therefore \(h(x)\) is not differentiable at \(x=0\).

Step 10: Check Differentiability for (D)

Differentiate \(j(x) = \sin (|x|)-|x|\): Here, both left and right derivatives at \(x=0\) match and equal \(1\), indicating that \(j(x)\) is differentiable at this point.

Key Concepts

Understanding Continuity

Continuity is a fundamental concept in calculus that ensures that a function behaves smoothly at a given point. For a function to be continuous at a point, such as at \(x=0\), it must meet the following criteria:

• The function value, \(f(0)\), must exist.
• The limit of the function as \(x\) approaches \(0\) from the left side, \(\lim_{{x \to 0^-}} f(x)\), and from the right side, \(\lim_{{x \to 0^+}} f(x)\), must exist and be equal.
• The limit of \(f(x)\) as \(x\) approaches \(0\) must be equal to \(f(0)\).

The step-by-step solution used these criteria to confirm the continuity of each function at \(x=0\). All the evaluated functions \(f(x), g(x), h(x),\) and \(j(x)\) met these conditions, showing that each is continuous at the specified point. Continuity serves as a prerequisite for checking differentiability, as a function cannot be differentiable where it is not continuous.

Exploring Absolute Value Functions

Absolute value functions are unique due to their piecewise nature, which can impact their differentiability. The absolute value \(\mid x\mid\) of \(x\) is defined as:

• \(x\) when \(x \geq 0\)
• -\(x\) when \(x < 0\)

This definition causes absolute value functions to create sharp turns or cusps, especially where \(x=0\). Such characteristics can make derivative analyses challenging, as derivatives handle changes in function values smoothly.
In the given exercise, absolute value factors within \(\cos(|x|)\) and \(\sin(|x|)\) functions created differing left- and right-hand derivatives at \(x=0\). Functions like \(g(x)\) and \(j(x)\) which incorporate \(\cos\) and \(\sin\) with absolute values show how studying these breakpoints is crucial for identifying differentiable cases.

Unraveling Derivatives

Derivatives represent how a function changes concerning its input, often described as the function's "instantaneous rate of change." For a function to be differentiable at a point, the left-hand derivative and the right-hand derivative at that point must match.
In mathematical terms, for a function \(f(x)\) to be differentiable at \(x=0\), the left-hand derivative \(f'(0^- )\) must equal the right-hand derivative \(f'(0^+)\). If they do not match, the function is not differentiable at that point.
In the problem, \(j(x) = \sin(|x|) - |x|\) was differentiable at \(x=0\) because both derivatives equaled \(1\). This showcases the importance of verifying derivatives from both sides to confirm differentiability, especially in functions containing absolute values, where non-smooth points frequently occur.