Problem 96 - 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

Option (D): Both are not differentiable at \(x=0\).

Step 1: Analyze Differentiability of f(x) at x=0

First, analyze the piecewise function \(f(x)\) at \(x=0\). \(f(x)\) switches from the constant value 3 for \(x<0\) to the linear function \(2x+1\) for \(x \geq 0\). Check that \(f(x)\) is continuous at \(x=0\). At \(x=0\), \(f(0) = 2(0) + 1 = 1\), which does not match \(f(x) = 3\) for \(x<0\). Thus, \(f(x)\) is not continuous at \(x=0\), and hence not differentiable.

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

Consider \(f(|x|)\), which means substituting \(|x|\) into \(f(x)\). For \(x<0\), \(|x| = -x\), substituting gives \(2(-x)+1 = 1-2x\), and for \(x\geq 0\), it remains \(2x+1\). For differentiability, note that \(f(|x|)\geq 0 = 1\), but at \(x<0\), \(\lim_{x \to 0^-} f(|x|) = 1\) and \(\lim_{x \to 0^+} f(|x|) = 1\). \(f(|x|)\) is continuous, but its left derivative and right derivative at \(x=0\) do not match \(\lim_{x \to 0^-} f'(|x|) = -2\), \(\lim_{x \to 0^+} f'(|x|) = 2\), so \(f(|x|)\) is not differentiable at \(x=0\).

Step 3: Conclusion

Neither \(f(x)\) nor \(f(|x|)\) is differentiable at \(x=0\). Hence, the correct option is (D).

Key Concepts

Piecewise Functions

A piecewise function is a type of function that has different expressions or rules for different intervals of the input values. These functions help us describe situations where a process or behavior changes at certain points or intervals.
\[ f(x) = \begin{cases} 3, & x<0 \ 2x+1, & x \geq 0 \end{cases} \]
In this example, for values of \(x\) less than zero, the function takes on the constant value 3. For values of \(x\) equal to or greater than zero, the function changes to a linear form, \(2x + 1\).

Piecewise functions are widely used in mathematics to model real-world situations where different rules apply to different domains. It's important to analyze each part of a piecewise function separately in order to understand its properties, such as continuity and differentiability.

Continuity

Continuity of a function at a point means that the function's value does not "jump" or have any gaps at that point. For a function to be continuous at a specific point, three conditions must be met:

• The function is defined at that point.
• The limit of the function as it approaches the point from either direction exists.
• The function's value at the point is the same as this limit.

In our exercise, we check the continuity of \(f(x)\) at \(x=0\).
At \(x=0\), the value of \(f(x)\) steps from a constant value 3 for \(x<0\) to \(2x+1\), giving \(f(0)=1\). Since \(f(0)eq 3\), \(f(x)\) is not continuous at \(x=0\).

Understanding continuity is crucial to analyzing more complex behavior of functions such as differentiability and integrability. It provides insight into how functions behave at certain points and across intervals.

Calculus

Calculus is a branch of mathematics that studies change. Differentiability is a central concept that deals with the rate at which a function changes. For a function to be differentiable at a point, it must have a well-defined and unique tangent, meaning the left-side derivative must match the right-side derivative at that point.
When evaluating differentiability for piecewise functions, such as \(f(x)\), one checks:

• Whether the function is continuous at the point in question.
• Whether the left-hand derivative equals the right-hand derivative at that point.

In the given problem, \(f(x)\) is not continuous at \(x=0\), so it is not differentiable. Moreover, \(f(|x|)\) is continuous, but the left and right derivatives do not match.

Understanding differentiability in calculus not only requires knowing about continuity but also involves limits and the precise definition of derivatives. This understanding is key when working with functions to pinpoint where and how they change, helping solve real-world problems and mathematical puzzles.