Problem 18 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\sum_{k=0}^{n} a_{k}|x-1|^{k}\), where \(a_{i} \in R\) then (A) \(f(x)\) is continuous at \(x=1\) for all \(a_{k} \in R\) (B) \(f(x)\) is differentiable at \(x=1\) for all \(a_{k} \in R\) (C) \(f(x)\) is differentiable at \(x=1\), provided \(a_{2 k+1}=0\) (D) \(f(x)\) is continuous at \(x=1\), provided \(a_{2 k}=0\)

Step-by-Step Solution

Answer

(A) and (C) are correct: (A) continuous for all real \( a_k \); (C) differentiable if odd powers vanish.

Step 1: Identify the Function Behavior at x = 1

To determine the continuity and differentiability of the function \( f(x) = \sum_{k=0}^{n} a_{k}|x-1|^{k} \) at \( x = 1 \), evaluate \( f(1) \). The function simplifies to \( f(1) = \sum_{k=0}^{n} a_{k} \cdot 0^{k} = a_0 \). Since \( f(1) \) equals \( a_0 \), which is a real number, the function is continuous at \( x=1 \) provided the expression simplifies properly around this point.

Step 2: Check Continuity Conditions

For \( f(x) \) to be continuous at \( x=1 \), both left and right limits as \( x \to 1 \) must equal \( f(1) \). As \( x \to 1 \), \( |x-1|^k \to 0 \) for k > 0, thus the limit from both sides approaches \( a_0 \). Therefore, \( f(x) \) is continuous at \( x=1 \) for all \( a_k \in \mathbb{R} \), confirming option (A).

Step 3: Examine Differentiability at x = 1

To determine differentiability at \( x = 1 \), compute \( f'(x) \) and check if it is continuous at this point. The function involving \( |x-1| \) complicates differentiability where the derivative form differs on the interval split by \( x=1 \), generally making it not differentiable unless odd k terms vanish.

Step 4: Conditions for Differentiability

If \( a_{2k+1}=0 \), all terms with odd powers vanish, representing a symmetrical behavior around the discontinuity at \( x = 1 \). This condition allows differentiability since \( \\sum_{k=0}^{n} a_{2k}|x-1|^{2k} \) will have a smooth derivative at the origin. Thus, option (C) is valid when odd terms are zero.

Key Concepts

Continuity at a point

Continuity at a point is a fundamental concept in calculus that tells us whether a function is smooth and unbroken at a specific point. For a function to be continuous at a point \( c \), three conditions must be met:

• The function \( f(x) \) must be defined at \( x = c \), meaning \( f(c) \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( c \) must exist.
• The limit and the function's value at the point must be equal: \( \lim_{{x \to c}} f(x) = f(c) \).

In the given problem, the function \( f(x) = \sum_{k=0}^{n} a_{k}|x-1|^{k} \) is being evaluated at \( x = 1 \). At this point, the expression simplifies, with most terms becoming zero. Only \( a_0 \) remains, confirming that \( f(1) = a_0 \). This simplification demonstrates that the function is continuous at \( x = 1 \), as both the left-hand and right-hand limits equal \( a_0 \), for any real number coefficients \( a_k \). That's why the function is continuous at \( x=1 \).

Differentiability at a point

Differentiability at a point indicates that a function has a definite slope or tangent at that point. For a function to be differentiable at a certain point \( x = c \), it must first be continuous at this point. Additionally, the derivative \( f'(x) \) must exist there. The derivative gives us the rate at which a function changes at a specific point.

In the problem's context, figuring out differentiability at \( x=1 \) involves the function with absolute terms \( |x-1| \). These absolute value terms are tricky because they can cause sharp turns in the graph. This sharpness generally disrupts differentiability, as a function might not have a well-defined tangent at these points.

However, when coefficients for odd powers of \( |x-1| \) become zero (\( a_{2k+1}=0 \)), the odd terms in the series vanish. This adjustment removes the sharp turning at \( x=1 \), permitting the function to have a smooth derivative there. Hence, the series becomes a sum of even powers, avoiding the sharp changes and making the function differentiable at \( x=1 \), which fulfills option (C).

Piecewise functions

Piecewise functions are functions defined by different expressions based on the input's value. They often arise when a function has different rules or patterns over different intervals of the domain. A classic example of a piecewise function is the absolute value function, which fits the structure \( |x| = \left\{\begin{array}{ll} -x, & \text{if } x < 0 \ x, & \text{if } x \geq 0 \end{array}\right. \).

The given problem's function \( f(x) = \sum_{k=0}^{n} a_{k}|x-1|^{k} \) behaves like a piecewise function because \(|x-1|\) acts differently depending on whether \( x-1 \) is positive or negative. This behavior splits the function's domain and can introduce potential discontinuities or challenges in differentiability.

Despite its piecewise nature, the function in the exercise is continuous at \( x=1 \) due to the way the sum simplifies. Yet, without specific conditions such as zeroing odd coefficients, differentiability at \( x=1 \) is not guaranteed. Understanding these transitions helps you grasp how piecewise functions operate and the specific conditions under which they can behave smoothly.