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

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

Both functions are not differentiable at \(x=0\). Option (D) is correct.

Step 1 - Recall the Definition

A function is differentiable at a point if its derivative exists at that point. This requires the function to be continuous at that point and both the left-hand derivative (LHD) and right-hand derivative (RHD) must exist and be equal.

Step 2 - Evaluate Differentiability of f(x) at x=0

Assess the continuity of \(f(x)\) at \(x=0\):- \(f(x) = 3\) when \(x < 0\)- \(f(x) = 2x + 1\) when \(x \geq 0\).At \(x=0\), \(f(0) = 2(0) + 1 = 1\).Since the left-hand limit is \(3\) and the right-hand limit is \(1\), \(f(x)\) is not continuous at \(x=0\), and therefore, not differentiable.

Step 3 - Evaluate Differentiability of f(|x|) at x=0

For \(f(|x|)\), when \(x<0\), \(f(|x|) = f(-x) = 2(-x)+1 = -2x + 1\).Find LHD and RHD at \(x=0\):- LHD: \(\lim_{h \to 0^-} \frac{f(|h|) - f(0)}{h} = \lim_{h \to 0^-} \frac{((-2h+1) - 1)}{h} = \lim_{h \to 0^-} \frac{-2h}{h} = -2\)- RHD: \(\lim_{h \to 0^+} \frac{f(|h|) - f(0)}{h} = \lim_{h \to 0^+} \frac{(2h+1 - 1)}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2\)Since LHD \(eq\) RHD, \(f(|x|)\) is not differentiable at \(x=0\).

Step 4: Conclusion

Both \(f(x)\) and \(f(|x|)\) are not differentiable at \(x=0\). Therefore, the correct answer is option (D).

Key Concepts

Piecewise Functions

Piecewise functions are fascinating mathematical entities because they allow us to describe a function with several sub-functions, each with its own domain. This means that the function behaves differently depending on the value of the input. A great example is the function given in the exercise:

• For values where \( x < 0 \), the function is constant at \( f(x) = 3 \).
• When \( x \geq 0 \), the function switches to \( f(x) = 2x + 1 \).

This dual nature allows piecewise functions to model situations where a rule changes its form after a certain threshold. While piecewise functions can be continuous and smooth, they often present challenges at the points where these sub-functions meet. At these junctions, which we often need to examine carefully, the function might not be smooth or even continuous. Differentiability of the function at these key points hinges on the continuity and smooth transition between these pieces.

Left-hand Derivative

The concept of the left-hand derivative (LHD) allows us to understand how a function approaches a point from the left, effectively capturing the slope of the function as we approach that point from negative values. For the function \( f(x) \) and its use with the absolute function \( f(|x|) \), it's crucial to compute LHD especially around critical points like \( x = 0 \).

• The LHD is defined as the following limit: \( \lim_{h \to 0^-} \frac{f(x+h) - f(x)}{h} \).
• For \( f(x) \) and \( f(|x|) \) at \( x=0 \), it's important to check what the function approaches as \( x \) moves in from the negative side.

In the case of \( f(|x|) \), for \( x < 0 \), since the function expresses \( f(|x|) = -2x + 1 \), the LHD at \( x=0 \) equals \( -2 \). This provides insight into the behavior of the function as it approaches from the left side.

Right-hand Derivative

Similarly, the right-hand derivative (RHD) lets us discern how a function behaves as it approaches the point from the right, capturing the slope from the positive side. Determining this is just as crucial as finding the LHD for differentiability.

• The RHD is given by the limit: \( \lim_{h \to 0^+} \frac{f(x+h) - f(x)}{h} \).
• For \( f(x) \) and \( f(|x|) \) at \( x=0 \), we observe the function as \( x \) nears zero from the positive side.

For the function \( f(|x|) \) in our exercise, the behavior changes to \( f(|x|) = 2x + 1 \) when \( x \geq 0 \), and thus the RHD at \( x=0 \) results in a value of \( 2 \). By comparing these derivatives, if they don't align, the function isn't considered differentiable at that point, marking the contrast in behavior from both sides.