Problem 1 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\left\\{\begin{array}{l}3, x<0 \\ 2 x+1, x \geq 0\end{array}\right.\), then (A) both \(f(x)\) and \(f(|x|)\) are differentiable at \(x=0\) (B) \(f(x)\) is differentiable but \(f(|x|)\) is not differentiable at \(x=0\) (C) \(f(|x|)\) is differentiable but \(f(x)\) is not differentiable at \(x=0\) (D) both \(f(x)\) and \(f(|x|)\) are not differentiable at \(x=0\)

Step-by-Step Solution

Answer

D. Both \(f(x)\) and \(f(|x|)\) are not differentiable at \(x=0\).

Step 1: Evaluate f(x) at x=0

We start by determining the function value at \(x = 0\) for \(f(x)\). Since \(f(x)\) is defined as \(f(x) = 2x + 1\) when \(x \geq 0\), we can evaluate \(f(0) = 2 \times 0 + 1 = 1\).

Step 2: Check Left-Hand Derivative of f(x)

Calculate the limit of \(f(x)\) as \(x\) approaches 0 from the left. For \(x < 0\), \(f(x) = 3\), thus \(\lim_{{x \to 0^-}} f(x) = 3\).

Step 3: Check Right-Hand Derivative of f(x)

Calculate the limit of \(f(x)\) as \(x\) approaches 0 from the right. For \(x \geq 0\), \(f(x) = 2x + 1\), therefore, \(\lim_{{x \to 0^+}} f(x) = 1\).

Step 4: Compare Limits and Continuity

Since \(\lim_{{x \to 0^-}} f(x) eq \lim_{{x \to 0^+}} f(x)\), \(f(x)\) is not continuous at \(x = 0\). A function must be continuous to be differentiable, so \(f(x)\) is not differentiable at \(x = 0\).

Step 5: Evaluate f(|x|) at x=0

We examine \(f(|x|)\), which changes the function's domain behavior for negative \(x\). For \(|x| < 0\), \(f(|x|) = 3\) (though \(|x|\) cannot be negative), while \(|x| \geq 0\) simplifies to \(f(x) = 2|x| + 1\). Thus, at \(x = 0\), \(f(|x|) = 1\).

Step 6: Check Differentiability of f(|x|)

The function \(f(|x|) = 2|x| + 1\) is simplified to \(f(x) = 1\) when evaluated at \(|x| = 0\). Then, \(\lim_{{x \to 0^-}} f(|x|) = \lim_{{x \to 0^+}} f(|x|) = 1\). Since both limits exist and equal the function value, we calculate the derivative: \(f'(|x|) = \pm 2\), which leads to non-differentiability as one limit approaches a negative slope, and the other a positive slope.

Step 7: Conclusion

Since \(f(x)\) is not continuous and thus not differentiable at \(x = 0\), and \(f(|x|)\) provides non-matching side derivatives, both are not differentiable at \(x = 0\).

Key Concepts

Piecewise Functions

Piecewise functions are interesting mathematical constructs. They are defined by different expressions depending on the input values' conditions. For instance, the function we're discussing today is defined as follows: for any input less than zero, it returns three, and for any input greater than or equal to zero, it returns the value of the expression \(2x + 1\).
These distinct rules allow piecewise functions to exhibit different behaviors over their domain, making their graph appear as separate pieces stitched together. This uniqueness often leads to complexity when analyzing properties like continuity and differentiability at certain points, especially where the transition between the pieces occurs.

Continuity

Continuity is a crucial concept in calculus, meaning that a function does not have any abrupt changes or jumps anywhere in its domain. Mathematically, a function \(f(x)\) is continuous at a point \(x = a\) if three conditions are met:

• \(f(a)\) is defined
• \(\lim_{{x \to a^-}} f(x) = \lim_{{x \to a^+}} f(x)\)
• \(\lim_{{x \to a}} f(x) = f(a)\)

If any of these conditions fail, the function has a discontinuity at that point. In our exercise, we check for continuity at \(x = 0\) and find that the left-hand limit and right-hand limit don't match. This reveals a jump discontinuity at this point, leading to the conclusion that \(f(x)\) is not continuous here.

Left-Hand Derivative

Derivatives tell us how fast a function changes at any given point, offering a sense of its slope or rate of change. The left-hand derivative focuses specifically on the behavior of the function as the input approaches from the left side.
To calculate the left-hand derivative at \(x = 0\) for our piecewise function, we consider the definition for \(x < 0\), which is a constant value 3. This means that as we approach 0 from the left, the function value remains constant, leading to a left-hand derivative of zero, indicating no change. However, the difference between this constant and the value defined for \(x \geq 0\) creates a problem for differentiability.

Right-Hand Derivative

Conversely, the right-hand derivative looks at how the function behaves as it approaches a point from the right side. In this instance, as \(x\) approaches 0 from the right, our function behaves according to \(2x + 1\). To find the right-hand derivative at \(x = 0\), evaluate the slope of this part of the function. Differentiating \(2x + 1\) gives us 2 as the rate of change.
Thus, the function has a right-hand slope of 2 just after \(x = 0\). The difference in these slopes from the left-hand and right-hand approach indicates that the slopes are not equal, leading us to the conclusion that the function is not differentiable at \(x = 0\). This mismatch in derivative values is a common cause for a failure in differentiability in piecewise functions.