Problem 27 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=a[x]+b e^{|x|}+c|x|^{2}\), where \(a, b\) and \(c\) are real constants. If \(f(x)\) is differentiable at \(x=0\), then (A) \(b=0, c=0, a \in R\) (B) \(a=0, c=0, b \in R\) (C) \(a=0, b=0, c \in R\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (A) \( b=0, c=0, a \in R \).

Step 1: Understanding differentiability

A function is differentiable at a point if it is continuous there, and if the derivative exists there. For this problem, we need to ensure that the derivative at 0 exists, meaning the left-hand and right-hand derivatives at 0 are equal.

Step 2: Analyze continuity at x=0

For the function to be continuous at 0, the limit of the function as x approaches 0 must be equal to the function's value at 0. Evaluate the expression at x=0, which is \( f(0) = a[0] + b e^{|0|} + c|0|^2 = a \) as both exponential and absolute squared terms vanish.

Step 3: Calculate derivatives and examine at x=0

The derivative from the right side is given by differentiating each component at x>0: \( f'(x) = rac{d}{dx}[a[x]+b e^{x}+c x^{2}] = b e^{x} + 2c x \). At x=0, this becomes \( b \). Similarly, the derivative from the left side \( x<0 \) yields \( rac{d}{dx}[a[x]+b e^{-x}+c x^{2}] = -b e^{-x} + 2c x \), which at x=0 also simplifies to \( -b \).

Step 4: Equate left-hand and right-hand derivatives

For differentiability at zero, ensure both \( b = -b \), leading to \( 2b = 0 \) and thus \( b = 0 \).

Step 5: Apply conditions for other constants

With b=0, the derivative conditions are now fully dependent on \( 2c = 0 \), leading to \( c = 0 \), and a can remain any real number since it does not affect differentiability at zero as it doesn't appear in the derivative expressions.

Key Concepts

Continuity

When talking about a function's continuity, we're referring to the idea that the function exhibits no abrupt changes, jumps, or breaks at a certain point. For a function to be continuous at a point, the following must happen:

• The function must be defined at that point.
• The limit of the function as it approaches that point from the left and the right must exist.
• Finally, the limit of the function at that point must equal the value of the function at that point.

In our exercise, we have the function \(f(x) = a[x] + b e^{|x|} + c|x|^2\). To verify if it is continuous at \(x=0\), we evaluate it at this point: \(f(0) = a\). _Why just \(a\)?_ Because both the exponential and squared terms vanish, given that \(e^{|0|} = 1\) and \(|0|^2 = 0\). This means \(f(x)\) is continuous at \(x=0\) if \(a\) is simply equal to itself, which is always true. This sets a solid foundation for checking differentiability here.

Derivative

The derivative of a function at a point is the 'slope' or 'rate of change' at that point. In formal terms, it's the limit of the average rate of change of the function over a small interval as the interval size approaches zero.
For differentiability at a point, a function must first be continuous at that point. Moreover, the left-hand derivative (approaching from the left) should equal the right-hand derivative (approaching from the right).
Let's examine this condition with our function \(f(x)\): \

• For \(x > 0\): \(f'(x) = be^{x} + 2cx\). Evaluated at 0, this gives \(f'(0^+) = b\).
• For \(x < 0\): \(f'(x) = -be^{-x} + 2cx\). Evaluated at 0, this yields \(f'(0^-) = -b\).

To make sure \(f(x)\) is differentiable at \(x=0\), we set these derivatives equal: \(b = -b\). Solving this gives \(b = 0\). Since \(b = 0\), the condition \(2c = 0\) follows, ensuring \(c = 0\) as well. Thus, differentiability at \(x=0\) requires both \(b\) and \(c\) to be zero.

Real Constants

In this exercise, \(a\), \(b\), and \(c\) are referred to as real constants. What do 'real constants' mean? These represent fixed values from the set of real numbers, as opposed to variables, which can change their values or take on any number from the set.

• In our function, \(a[x]\), \(b e^{|x|}\), and \(c |x|^2\) involve these constants which influence the function's shape and behavior.
• When terms like \(b\) and \(c\) were determined to be zero for the function to be differentiable at \(x=0\), they ceased to impact the derivative conditions since the derivative depends on these constant terms.
• \(a\) on the other hand, being a real constant, doesn't affect differentiation because when differentiating, constant terms vanish.

Therefore, as a conclusion, only \(a\) remains free within our solution, demonstrating how different constant terms can independently affect the criteria for a function being differentiable or continuous at particular points.